Putnam All Questions
-
Upload
gaurav-tiwari -
Category
Documents
-
view
217 -
download
0
Transcript of Putnam All Questions
-
7/26/2019 Putnam All Questions
1/109
The Forty-Sixth Annual William Lowell Putnam Competition
Saturday, December 7, 1985
A1 Determine, with proof, the number of ordered triples
of sets which have the property that
(i)
,and
(ii)
.
Express your answer in the form
, where
are nonnegative integers.
A2 Let be an acute triangle. Inscribe a rectangle in
with one side along a side of . Then inscribe a rectan-
gle in the triangle formed by the side of
opposite
the side on the boundary of , and the other two sides of
, with one side along the side of
. For any polygon
, let
denote the area of
. Find the maximumvalue, or show that no maximum exists, of
,
where ranges over all triangles and
over all rect-
angles as above.
A3 Let
be a real number. For each integer
, define
a sequence
,
by the condition
Evaluate
.
A4 Define a sequence
by
and
for
. Which integers between 00 and 99 inclusive oc-
cur as the last two digits in the decimal expansion of
infinitely many ?
A5 Let
. For
which integers ,
is
?
A6 If
is a polynomial with
real coefficients , then set
Let
. Find, with proof, a polyno-
mial
with real coefficients such that
(i)
, and
(ii)
for every integer
.
B1 Let be the smallest positive integer for which there ex-ist distinct integers
such that the
polynomial
has exactly nonzero coefficients. Find, with proof, a
set of integers
for which this min-
imum is achieved.
B2 Define polynomials
for
by
,
for
, and
for
. Find, with proof, the explicit factorization of
into powers of distinct primes.
B3 Let
......
.... . .
be a doubly infinite array of positive integers, and sup-
pose each positive integer appears exactly eight times
in the array. Prove that
for some pair ofpositive integers
.
B4 Let be the unit circle
. A point is chosen
randomly on the circumference and another point
is chosen randomly from the interior of (these points
are chosen independently and uniformly over their do-
mains). Let be the rectangle with sides parallel to the
and
-axes with diagonal . What is the probability
that no point of lies outside of ?
B5 Evaluate
. You may assume
that
.
B6 Let be a finite set of real
matrices
,
, which form a group under matrix multi-
plication. Suppose that
, where
denotes the trace of the matrix
. Prove that
is the
zero matrix.
-
7/26/2019 Putnam All Questions
2/109
The Forty-Seventh Annual William Lowell Putnam Competition
Saturday, December 6, 1986
A1 Find, with explanation, the maximum value off(x) =x3 3xon the set of all real numbersxsatisfyingx4 +36
13x2.
A2 What is the units (i.e., rightmost) digit of
1020000
10100 + 3
?
A3 Evaluate
n=0Arccot(n2+n+1), where Arccot tfor
t 0denotes the number in the interval0< /2withcot = t.
A4 A transversalof annnmatrixAconsists ofnentriesofA, no two in the same row or column. Let f(n) bethe number ofnn matricesAsatisfying the followingtwo conditions:
(a) Each entryi,j ofAis in the set {1, 0, 1}.(b) The sum of the n entries of a transversal is the
same for all transversals ofA.
An example of such a matrixA is
A=
1 0 10 1 0
0 1 0
.
Determine with proof a formula forf(n)of the form
f(n) =a1bn1 +a2bn2 +a3bn3 +a4,
where theais andbis are rational numbers.
A5 Suppose f1(x), f2(x), . . . , f n(x)are functions ofnrealvariablesx = (x1, . . . , xn) with continuous second-order partial derivatives everywhere on Rn. Suppose
further that there are constantscij such that
fixj
fjxi
=cij
for alli and j ,1 i n,1 j n. Prove that thereis a function g(x)on Rn such thatfi + g/xiis linearfor all i, 1 i n. (A linear function is one of theform
a0+a1x1+a2x2+ +anxn.)
A6 Let a1, a2, . . . , an be real numbers, and letb1, b2, . . . , bn be distinct positive integers. Sup-pose that there is a polynomial f(x) satisfying theidentity
(1 x)nf(x) = 1 +n
i=1
aixbi .
Find a simple expression (not involving any sums) fof(1) in terms ofb1, b2, . . . , bn andn (but independenofa1, a2, . . . , an).
B1 Inscribe a rectangle of base b and height h in a circlof radius one, and inscribe an isosceles triangle in th
region of the circle cut off by one base of the rectangl
(with that side as the base of the triangle). For whavalue ofh do the rectangle and triangle have the samarea?
B2 Prove that there are only a finite number of possibilitie
for the ordered tripleT = (x y, y z, z x), wherx,y,z are complex numbers satisfying the simultaneous equations
x(x 1) + 2yz = y(y 1) + 2zx = z(z 1) + 2xy,and list all such triplesT.
B3 Let consist of all polynomials inx with integer coefficients. Forf andg in and m a positive integer, lef g (mod m) mean that every coefficient offis an integral multiple ofm. Let n and p be positivintegers withp prime. Given thatf , g , h , rand s are i withrf+sg 1 (mod p) and f g h (mod p)prove that there exist F and G in with F f(modp),G g (mod p), andF G h (mod pn).
B4 For a positive real numberr, letG(r)be the minimumvalue of|r m2 + 2n2| for all integers m and nProve or disprove the assertion that limrG(r) exists and equals 0.
B5 Let f(x,y,z) = x2 + y2 + z2 + xyz . Lep(x,y,z), q(x,y,z), r(x,y,z) be polynomials witreal coefficients satisfying
f(p(x,y,z), q(x,y,z), r(x,y,z)) =f(x,y,z).
Prove or disprove the assertion that the sequencep,q,consists of some permutation ofx,y,z, where thnumber of minus signs is 0 or 2.
B6 SupposeA,B, C,D are n nmatrices with entries ia fieldF, satisfying the conditions thatABT andCDT
are symmetric and ADT BCT = I. HereI is thnnidentity matrix, and ifMis annnmatrix,MTis its transpose. Prove thatATD CTB= I.
-
7/26/2019 Putnam All Questions
3/109
The Forty-Eighth Annual William Lowell Putnam Competition
Saturday, December 5, 1987
A1 Curves
and are defined in the plane as fol-
lows:
Prove that
.
A2 The sequence of digits
is obtained by writing the positive integers in order. If
the
-th digit in this sequence occurs in the part ofthe sequence in which the -digit numbers are placed,
define
to be . For example,
because
the 100th digit enters the sequence in the placement of
the two-digit integer 55. Find, with proof,
.
A3 For all real
, the real-valued function
satis-
fies
(a) If
for all real
, must
for all
real
? Explain.
(b) If
for all real
, must
for all
real
? Explain.A4 Let
be a polynomial, with real coefficients, in three
variables and be a function of two variables such that
for all real
and such that
,
, and
. Also let
be complex numbers
with
and
. Find
.
A5 Let
Prove or disprove that there is a vector-valued function
with the following properties:
(i)
have continuous partial derivatives for
all
;
(ii)
for all
;
(iii)
.
A6 For each positive integer
, let
be the number of
zeroes in the base 3 representation of
. For which pos-
itive real numbers
does the series
converge?
B1 Evaluate
B2 Let
and be integers with
,
and
. Prove that
B3 Let be a field in which
. Show that the set
of solutions to the equation
with
and
in is given by
and
where runs through the elements of such that
.
B4 Let
and let
and
for
. For each of
and
,
prove that the limit exists and find it or prove that thelimit does not exist.
B5 Let be the
-dimensional vector
. Let
be a
matrix of complex numbers such that
whenever
, with complex
,
not all zero, then at least one of the
is not real. Prove
that for arbitrary real numbers
, there are
complex numbers
such that
...
...
(Note: if
is a matrix of complex numbers,
isthe matrix whose entries are the real parts of the entries
of
.)
B6 Let be the field of
elements, where is an odd
prime. Suppose is a set of
distinct nonzero
elements of with the property that for each
in
, exactly one of and
is in . Let
be the num-
ber of elements in the intersection
.
Prove that
is even.
-
7/26/2019 Putnam All Questions
4/109
The Forty-Ninth Annual William Lowell Putnam Competition
Saturday, December 3, 1988
A1 Let
be the region consisting of the points of thecartesian plane satisfying both
and
.
Sketch the region
and find its area.
A2 A not uncommon calculus mistake is to believe that the
product rule for derivatives says that . If
, determine, with proof, whether there exists
an open interval
and a nonzero function defined
on
such that this wrong product rule is true for
in
.
A3 Determine, with proof, the set of real numbers for
which
converges.
A4 (a) If every point of the plane is painted one of three
colors, do there necessarily exist two points of the
same color exactly one inch apart?
(b) What if three is replaced by nine?
A5 Prove that there exists auniquefunction from the set
of positive real numbers to
such that
and
for all
.
A6 If a linear transformation on an -dimensional vector
space has
eigenvectors such that any of them
are linearly independent, does it follow that is a scalar
multiple of the identity? Prove your answer.
B1 A composite (positive integer) is a product with
and not necessarily distinct integers in
.
Show that every composite is expressible as
, with
positive integers.
B2 Prove or disprove: If and are real numbers with
and
, then
.
B3 For every in the set
of positive inte-
gers, let
be the minimum value of
for all
nonnegative integers and
with
. Find, with
proof, the smallest positive real number with
for all
.
B4 Prove that if
is a convergent series of positive
real numbers, then so is
.
B5 For positive integers , let be the by
skew-symmetric matrix for which each entry in the first subdiagonals below the main diagonal is 1 and each
of the remaining entries below the main diagonal is -1.
Find, with proof, the rank of
. (According to one
definition, the rank of a matrix is the largest such that
there is a
submatrix with nonzero determinant.)
One may note that
B6 Prove that there exist an infinite number of ordered pairs
of integers such that for every positive integer
,
the number
is a triangular number if and only if
is a triangular number. (The triangular numbers are the
with in
.)
-
7/26/2019 Putnam All Questions
5/109
The Fiftieth Annual William Lowell Putnam Competition
Saturday, December 2, 1989
A1 How many primes among the positive integers, writtenas usual in base 10, are alternating 1s and 0s, begin-
ning and ending with 1?
A2 Evaluate
where and
are positive.
A3 Prove that if
then
(Here
is a complex number and
.)
A4 If is an irrational number,
, is there afinite game with an honest coin such that the probabil-
ity of one player winning the game is ? (An honest
coin is one for which the probability of heads and the
probability of tails are both
. A game is finite if with
probability 1 it must end in a finite number of moves.)
A5 Let be a positive integer and let
be a regular
-gon inscribed in the unit circle. Show that
there is a positive constant , independent of
, with
the following property. For any points inside
there
are two distinct vertices
and
of such that
Here
denotes the distance between the points
and
.
A6 Let
be a formal power series
with coefficients in the field of two elements. Let
if every block of zeros in the binary
expansion of has an even number
of zeros in the block
otherwise.
(For example,
because
and
because
) Prove that
B1 A dart, thrown at random, hits a square target. Assum-
ing that any two parts of the target of equal area are
equally likely to be hit, find the probability that the point
hit is nearer to the center than to any edge. Express your
answer in the form
, where
are inte-
gers.
B2 Let be a non-empty set with an associative opera-tion that is left and right cancellative (
implies
, and
implies
). Assume that for
every in
the set
is finite.
Must be a group?
B3 Let be a function on
, differentiable and satis-
fying
for
. Assume that
for
(so
that
tends rapidly to
as
increases). For a
non-negative integer, define
(sometimes called the th moment of
).
a) Express in terms of .
b) Prove that the sequence
always con-
verges, and that the limit is
only if
.
B4 Can a countably infinite set have an uncountable collec-
tion of non-empty subsets such that the intersection of
any two of them is finite?
B5 Label the vertices of a trapezoid (quadrilateral
with two parallel sides) inscribed in the unit circle
as
so that
is parallel to
and
are in counterclockwise order. Let
,
and
denote the lengths of the line segments
,
and , where E is the point of intersection of the di-
agonals of , and is the center of the circle. Deter-
mine the least upper bound of
over all such for
which
, and describe all cases, if any, in which it
is attained.
B6 Let
be a point chosen at random from
the -dimensional region defined by
Let be a continuous function on
with
. Set
and
. Show that
the expected value of the Riemann sum
is
, where
is a polynomial of degree ,
independent of , with
for
.
-
7/26/2019 Putnam All Questions
6/109
The 51st William Lowell Putnam Mathematical Competition
Saturday, December 8, 1990
A1 Let
T0= 2, T1= 3, T2= 6,
and forn3,Tn= (n + 4)Tn14nTn2+ (4n 8)Tn3.
The first few terms are
2, 3, 6, 14, 40, 152, 784, 5168, 40576.
Find, with proof, a formula for Tn of the form Tn =An+ Bn, where{An} and{Bn} are well-known se-quences.
A2 Is
2 the limit of a sequence of numbers of the form3
n 3m(n, m= 0, 1, 2, . . . )?A3 Prove that any convex pentagon whose vertices (no
three of which are collinear) have integer coordinates
must have area greater than or equal to 5/2.
A4 Consider a paper punch that can be centered at any pointof the plane and that, when operated, removes from the
plane precisely those points whose distance from the
center is irrational. How many punches are needed to
remove every point?
A5 IfA and B are square matrices of the same size suchthat ABAB = 0, does it follow that BABA = 0?
A6 If X is a finite set, let X denote the number of ele-ments inX. Call an ordered pair(S, T) of subsets of{1, 2, . . . , n} admissibleifs >|T| for eachsS, andt >|S|for eachtT. How many admissible orderedpairs of subsets of{1, 2, . . . , 10} are there? Prove youranswer.
B1 Find all real-valued continuously differentiable func-tionsfon the real line such that for allx,
(f(x))2 = x
0
[(f(t))2 + (f(t))2] dt + 1990.
B2 Prove that for |x|< 1, |z|> 1,
1 +
j=1
(1 + xj)Pj = 0,
wherePj is
(1 z)(1 zx)(1 zx2) (1 zxj1)(z x)(z x2)(z x3) (z xj) .
B3 Let S be a set of 2 2 integer matrices whose entries aij (1) are all squares of integers and, (2) satisfy aij 200. Show that ifShas more than 50387(= 154 152 15 + 2) elements, then it has two elements that commute.
B4 LetGbe a finite group of orderngenerated byaandbProve or disprove: there is a sequence
g1, g2, g3, . . . , g2n
such that
(1) every element ofG occurs exactly twice, and
(2) gi+1equalsgiaorgibfori= 1, 2, . . . , 2n. (Interpretg2n+1as g1.)
B5 Is there an infinite sequencea0, a1, a2, . . . of nonzerreal numbers such that for
n = 1, 2, 3, . . . the polyno
mial
pn(x) = a0+ a1x + a2x2 + + anxn
has exactlyn distinct real roots?
B6 LetSbe a nonempty closed bounded convex set in thplane. LetK be a line and t a positive number. LeL1 and L2 be support lines for Sparallel to K1, anletL be the line parallel toKand midway betweenLandL2. LetBS(K, t)be the band of points whose distance fromLis at most(t/2)w, wherew is the distanc
betweenL1and L2. What is the smallestt such that
SK
BS(K, t)=
for allS? (Kruns over all lines in the plane.)
-
7/26/2019 Putnam All Questions
7/109
The Fifty-Second William Lowell Putnam Mathematical Competition
Saturday, December 7, 1991
A1 A
rectangle has vertices as
and
. It rotates
clockwise about the point
. It
then rotates
clockwise about the point
, then
clockwise about the point
, and finally,
clockwise about the point
. (The side originally
on the -axis is now back on the
-axis.) Find the
area of the region above the -axis and below the curve
traced out by the point whose initial position is (1,1).
A2 Let and
be different
matrices with real en-
tries. If and , can be
invertible?
A3 Find all real polynomials
of degree
for
which there exist real numbers
such that
1.
and
2.
where
denotes the derivative of
.
A4 Does there exist an infinite sequence of closed
discs
in the plane, with centers
, respectively, such that
1. the have no limit point in the finite plane,
2. the sum of the areas of the
is finite, and3. every line in the plane intersects at least one of the
?
A5 Find the maximum value of
for .
A6 Let denote the number of sums of positive inte-
gers
which add up to with
Let
denote the number of
which
add up to , with
1.
2. each is in the sequence
de-fined by
,
, and
and
3. if
then every element in
appears at least once as a
.
Prove that
for each
.
(For example, because the relevant sums are
and
because
the relevant sums are
)
B1 For each integer
, let
, where
is the greatest integer with
. Define a sequence
by
and
for
.
For what positive integers is this sequence eventually
constant?
B2 Suppose and
are non-constant, differentiable, real-
valued functions defined on
. Furthermore,
suppose that for each pair of real numbers and
,
If
, prove that
for all
.
B3 Does there exist a real number such that, if and
are integers greater than , then an
rectangle
may be expressed as a union of
and
rect-
angles, any two of which intersect at most along their
boundaries?
B4 Suppose is an odd prime. Prove that
B5 Let be an odd prime and let
denote (the field of)
integers modulo . How many elements are in the set
B6 Let
and be positive numbers. Find the largest num-
ber , in terms of
and , such that
for all
with
and for all ,
.
(Note:
.)
-
7/26/2019 Putnam All Questions
8/109
The 53rd William Lowell Putnam Mathematical Competition
Saturday, December 5, 1992
A-1 Prove thatf(n) = 1nis the only integer-valued func-tion defined on the integers that satisfies the following
conditions.
(i) f(f(n)) = n, for all integersn;
(ii) f(f(n+ 2) + 2) = n for all integersn;
(iii) f(0) = 1.
A-2 DefineC()to be the coefficient ofx1992 in the powerseries aboutx = 0of(1 +x). Evaluate
10
C(y 1)
1992k=1
1
y+k
dy.
A-3 For a given positive integerm, find all triples (n,x,y)of positive integers, withnrelatively prime tom, whichsatisfy
(x2 +y2)m = (xy)n.
A-4 Letfbe an infinitely differentiable real-valued functiondefined on the real numbers. If
f
1
n
=
n2
n2 + 1, n= 1, 2, 3, . . . ,
compute the values of the derivatives f(k)(0), k =1, 2, 3, . . . .
A-5 For each positive integern, letan= 0(or 1) if the num-ber of 1s in the binary representation ofn is even (orodd), respectively. Show that there do not exist positiveintegersk and msuch that
ak+j =ak+m+j =ak+2m+j ,
for0 j m 1.
A-6 Four points are chosen at random on the surface of a
sphere. What is the probability that the center of the
sphere lies inside the tetrahedron whose vertices are atthe four points? (It is understood that each point is in-
dependently chosen relative to a uniform distribution on
the sphere.)
B-1 Let S be a set ofn distinct real numbers. LetAS bethe set of numbers that occur as averages of two distinct
elements ofS. For a givenn 2, what is the smallestpossible number of elements inAS?
B-2 For nonnegative integers n and k , defineQ(n, k)to bethe coefficient ofxk in the expansion of(1 +x+x2 +x3)n. Prove that
Q(n, k) =k
j=0
n
j
n
k 2j
,
whereab
is the standard binomial coefficient. (Re
minder: For integersaandbwitha 0,
ab
= a!
b!(ab)
for0 b a, with ab= 0otherwise.)
B-3 For any pair (x, y) of real numbers, a sequenc(an(x, y))n0 is defined as follows:
a0(x, y) = x,
an+1(x, y) = (an(x, y))
2 +y2
2 , forn 0.
Find the area of the region
{(x, y)|(an(x, y))n0 converges}.
B-4 Letp(x) be a nonzero polynomial of degree less tha1992 having no nonconstant factor in common with
x3 x. Let
d1992
dx1992
p(x)
x3 x
=
f(x)
g(x)
for polynomialsf(x) and g(x). Find the smallest possible degree off(x).
B-5 LetDndenote the value of the(n 1) (n 1)determinant
3 1 1 1 11 4 1 1 11 1 5 1 11 1 1 6 1...
......
.... . .
...
1 1 1 1 n+ 1
.
Is the setDnn!
n2
bounded?
B-6 Let M be a set of realn nmatrices such that
(i) I M, whereIis then nidentity matrix;
(ii) ifA Mand B M, then either AB MoAB M, but not both;
(iii) ifA M andB M, then eitherAB = BAoAB= BA;
(iv) ifA M andA=I, there is at least oneB Msuch thatAB = BA.
Prove that M contains at mostn2 matrices.
-
7/26/2019 Putnam All Questions
9/109
The 54th William Lowell Putnam Mathematical Competition
Saturday, December 4, 1993
A1 The horizontal liney = c intersects the curvey = 2x3x3 in the first quadrant as in the figure. Findc so thatthe areas of the two shaded regions are equal. [Figure
not included. The first region is bounded by they-axis,the line y = c and the curve; the other lies under thecurve and above the liney = cbetween their two pointsof intersection.]
A2 Let (xn)n0 be a sequence of nonzero real numberssuch that x2n xn1xn+1 = 1 for n = 1, 2, 3, . . . .Prove there exists a real number a such that xn+1 =axn xn1 for alln 1.
A3 Let Pn be the set of subsets of {1, 2, . . . , n}. Letc(n, m) be the number of functions f : Pn {1, 2, . . . , m} such that f(AB) = min{f(A), f(B)}.Prove that
c(n, m) =mj=1
jn.
A4 Let x1, x2, . . . , x19be positive integers each of which isless than or equal to 93. Lety1, y2, . . . , y93 be positiveintegers each of which is less than or equal to 19. Prove
that there exists a (nonempty) sum of somexis equalto a sum of someyjs.
A5 Show that
10100
x2 xx3 3x+ 1
2dx+
111
1
101
x2 x
x3 3x+ 1
2dx+
1110
101
100
x2 x
x3 3x+ 1
2dx
is a rational number.
A6 The infinite sequence of 2s and 3s
2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3,
3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, . . .
has the property that, if one forms a second sequencethat records the number of 3s between successive 2s,
the result is identical to the given sequence. Show that
there exists a real numberrsuch that, for anyn, thenthterm of the sequence is 2 if and only ifn = 1 + rmfor some nonnegative integer m. (Note:x denotes thelargest integer less than or equal tox.)
B1 Find the smallest positive integern such that for everintegermwith0< m < 1993, there exists an integerkfor which
m
1993 N.B4 Forn 1, letdn be the greatest common divisor of th
entries ofAn I, where
A=
3 24 3
and I=
1 00 1
.
Show thatlimn dn= .
B5 For any real number , define the function f(x) =x. Letnbe a positive integer. Show that there existan such that for1 k n,
fk(n2) =n2 k= fk(n2).
B6 For any integern, set
na= 101a 100 2a.
Show that for0 a,b,c,d 99,na+ nb nc+ n(mod 10100)implies {a, b} = {c, d}.
-
7/26/2019 Putnam All Questions
11/109
The Fifty-Sixth William Lowell Putnam Mathematical Competition
Saturday, December 2, 1995
A1 Let
be a set of real numbers which is closed under
multiplication (that is, if and
are in
, then so is ).
Let and be disjoint subsets of
whose union is
. Given that the product of any three(not necessarilydistinct) elements of is in and that the product of
any three elements of is in
, show that at least one
of the two subsets
is closed under multiplication.
A2 For what pairs of positive real numbers does the
improper integral
converge?
A3 The number has nine (not necessarily dis-
tinct) decimal digits. The number
is such
that each of the nine 9-digit numbers formed by replac-ing just one of the digits
is
by the corre-
sponding digit (
) is divisible by 7. The
number
is related to
is the same
way: that is, each of the nine numbers formed by replac-
ing one of the
by the corresponding
is divisible by
7. Show that, for each ,
is divisible by 7. [For
example, if
, then may be 2
or 9, since
and
are multiples of
7.]
A4 Suppose we have a necklace of beads. Each bead is
labeled with an integer and the sum of all these labels
is
. Prove that we can cut the necklace to form a
string whose consecutive labels
satisfy
for
A5 Let
be differentiable (real-valued) func-
tions of a single variable which satisfy
..
.
..
.
for some constants
. Suppose that for all ,
as
. Are the functions
necessarily linearly dependent?
A6 Suppose that each of people writes down the numbers
1,2,3 in random order in one column of a
matrix,
with all orders equally likely and with the orders for
different columns independent of each other. Let the
row sums
of the resulting matrix be rearranged
(if necessary) so that . Show that for some
, it is at least four times as likely that both
and
as that .
B1 For a partition of
, let
be
the number of elements in the part containing
. Prove
that for any two partitions and
, there are two dis-
tinct numbers
and in such
that
and
. [A partition of
a set
is a collection of disjoint subsets (parts) whose
union is
.]
B2 An ellipse, whose semi-axes have lengths and , rolls
without slipping on the curve . How are
related, given that the ellipse completes one rev-
olution when it traverses one period of the curve?
B3 To each positive integer with
decimal digits, we as-
sociate the determinant of the matrix obtained by writ-
ing the digits in order across the rows. For example, for
, to the integer 8617 we associate
. Find, as a function of , the sum of all the determi-
nants associated with
-digit integers. (Leading digits
are assumed to be nonzero; for example, for ,
there are 9000 determinants.)
B4 Evaluate
Express your answer in the form
, where
are integers.
B5 A game starts with four heaps of beans, containing 3,4,5
and 6 beans. The two players move alternately. A move
consists of takingeither
a) one bean from a heap, provided at least two beans
are left behind in that heap, or
b) a complete heap of two or three beans.
The player who takes the last heap wins. To win thegame, do you want to move first or second? Give a
winning strategy.
B6 For a positive real number , define
Prove that cannot be expressed as the dis-
joint union of three sets
and
. [As
usual,
is the greatest integer
.]
-
7/26/2019 Putnam All Questions
12/109
Solutions to the Fifty-Sixth William Lowell Putnam Mathematical Competition
Saturday, December 2, 1995
Kiran Kedlaya
A1 Suppose on the contrary that there exist
with
and
with
. Then
while
, contradiction.
A2 The integral converges iff . The easiest proof
uses big-O notation and the fact that
for
. (Here
means
bounded by a constant times
.)
So
hence
and similarly
Hence the integral were looking at is
The term
is bounded by a constant times
, whose integral converges. Thus we only have todecide whether
converges. But
has divergent integral, so we get convergence if and
only if (in which case the integral telescopes any-
way).
A3 Let and
be the numbers
and
, re-
spectively. We are given that
and
for
. Sum the first relation over
and we
get
, or
.
Now add the first and second relations for any particu-
lar value of and we get
. But we know
is divisible by 7, and 10
is coprime to 7, so
.
A4 Let
, so that
. These form a cyclic sequence that doesnt change
when you rotate the necklace, except that the entire se-
quence gets translated by a constant. In particular, it
makes sense to choose
for which is maximum and
make that one
; this way
for all , which gives
, but the right side may be
replaced by
since the left side is an integer.
A5 Everyone (presumably) knows that the set of solutions
of a system of linear first-order differential equations
with constant coefficients is
-dimensional, with ba-
sis vectors of the form
(i.e. a function times
a constant vector), where the
are linearly indepen-
dent. In particular, our solution
can be written as
.
Choose a vector
orthogonal to
but not to
. Since
as
, the same is true of
;
but that is simply
. In other words, if
, then
must also go to 0.
However, it is easy to exhibit a solution which does
not go to 0. The sum of the eigenvalues of the matrix
, also known as the trace of
, being the sum
of the diagonal entries of
, is nonnegative, so
has
an eigenvalue with nonnegative real part, and a cor-
responding eigenvector
. Then
is a solution that
does not go to 0. (If is not real, add this solution to
its complex conjugate to get a real solution, which still
doesnt go to 0.)
Hence one of the
, say
, is zero, in which case
for all
.
A6 View this as a random walk/Markov process with states
the triples of integers with sum 0, correspond-
ing to the difference between the first, second and third
rows with their average (twice the number of columns).
Adding a new column adds on a random permutation
of the vector
. I prefer to identify the triple
with the point
in
the plane, where
is a cube root of unity. Then adding
a new column corresponds to moving to one of the six
neighbors of the current position in a triangular lattice.
What wed like to argue is that for large enough
, the
ratio of the probabilities of being in any two particular
states goes to 1. Then in fact, well see that eventually,about six times as many matrices have
than
. This is a pain to prove, though,
and in fact is way more than we actually need.
Let and
be the probability that we are at the
origin, or at a particular point adjacent to the origin,
respectively. Then
. (In fact,
is
times the sum of the probabilities of being at each
neighbor of the origin at time
, but these are all
.)
-
7/26/2019 Putnam All Questions
13/109
So the desired result, which is that
for
some large
, is equivalent to
.
Suppose on the contrary that this is not the case; then
for some constant
. However, if
, the probability that we chose each of the six
types of moves
times is already
,
which by Stirlings approximation is asymptotic to a
constant times
. This term alone is bigger than
, so we must have
for some
. (In fact, we must have
for any
.)
B1 For a given , no more than three different values of
are possible (four would require one part each of
size at least 1,2,3,4, and thats already more than 9 el-
ements). If no such
exist, each pair
occurs for at most 1 element of
, and since there are
only
possible pairs, each must occur exactly once.
In particular, each value of
must occur 3 times.
However, clearly any given value of
occurs
times, where is the number of distinct partitions of
that size. Thus
can occur 3 times only if it equals
1 or 3, but we have three distinct values for which it
occurs, contradiction.
B2 For those who havent taken enough physics, rolling
without slipping means that the perimeter of the ellipse
and the curve pass at the same rate, so all were saying
is that the perimeter of the ellipse equals the length of
one period of the sine curve. So set up the integrals:
Let
in the second integral and write 1 as
and you get
Since the left side is increasing as a function of , we
have equality if and only if
.
B3 For
we obviously get 45, while for
the
answer is 0 because it both changes sign (because de-terminants are alternating) and remains unchanged (by
symmetry) when you switch any two rows other than
the first one. So only
is left. By the multilin-
earity of the determinant, the answer is the determinant
of the matrix whose first (resp. second) row is the sum
of all possible first (resp. second) rows. There are 90
first rows whose sum is the vector
, and 100
second rows whose sum is
. Thus the answer
is
B4 The infinite continued fraction is defined as the limit
of the sequence
.
Notice that the sequence is strictly decreasing (by in-
duction) and thus indeed has a limit , which satisfies
, or rewriting,
.
Moreover, we want the greater of the two roots.
Now how to compute the eighth root of ? Notice that
if
satisfies the quadratic
, then we
have
Clearly, then, the positive square roots of the quadratic
satisfy the quadratic
. Thus we compute that
is the greater
root of
,
is the greater root
of
, and
is the greater root of
, otherwise known as
.
B5 This problem is dumb if you know the Sprague-
Grundy theory of normal impartial games (see Conway,Berlekamp and Guy,Winning Ways, for details). Ill de-
scribe how it applies here. To each position you assign
a nim-value as follows. A position with no moves (in
which case the person to move has just lost) takes value
0. Any other position is assigned the smallest number
not assigned to a valid move from that position.
For a single pile, one sees that an empty pile has value
0, a pile of 2 has value 1, a pile of 3 has value 2, a pile
of 4 has value 0, a pile of 5 has value 1, and a pile of 6
has value 0.
You add piles just like in standard Nim: the nim-value
of the composite of two games (where at every turn you
pick a game and make a move there) is the base 2 ad-dition without carries (i.e. exclusive OR) of the nim-
values of the constituents. So our starting position, with
piles of 3, 4, 5, 6, has nim-value
.
A position is a win for the player to move if and only if
it has a nonzero value, in which case the winning strat-
egy is to always move to a 0 position. (This is always
possible from a nonzero position and never from a zero
position, which is precisely the condition that defines
the set of winning positions.) In this case, the winning
move is to reduce the pile of 3 down to 2, and you can
easily describe the entire strategy if you so desire.
B6 Obviously
have to be greater than 1, and no twocan both be rational, so without loss of generality as-
sume that and
are irrational. Let
denote the fractional part of
. Then
if and
only if
. In particular, this
means that
contains
elements, and similarly. Hence for every integer
,
2
-
7/26/2019 Putnam All Questions
14/109
Dividing through by
and taking the limit as
shows that
. That in turn implies
that for all
,
Our desired contradiction is equivalent to showing that
the left side actually takes the value 1 for some
.
Since the left side is an integer, it suffices to show that
for some
.
A result in ergodic theory (the two-dimensional version
of the Weil equidistribution theorem) states that if
are linearly independent over the rationals, then the set
of points
is dense (and in fact equidis-
tributed) in the unit square. In particular, our claim def-
initely holds unless
for some integers
.
On the other hand, suppose that such a relation
does hold. Since and
are irrational, by the
one-dimensional Weil theorem, the set of points
is dense in the set of
in the
unit square such that
is an integer. It is simpleenough to show that this set meets the region
unless
is an integer, and that
would imply that
, a quantity between 0 and
1, is an integer. We have our desired contradiction.
3
-
7/26/2019 Putnam All Questions
15/109
The Fifty-Seventh William Lowell Putnam Mathematical Competition
Saturday, December 7, 1996
A1 Find the least number
such that for any two squares of
combined area 1, a rectangle of area
exists such that
the two squares can be packed in the rectangle (without
interior overlap). You may assume that the sides of thesquares are parallel to the sides of the rectangle.
A2 Let
and
be circles whose centers are 10 units
apart, and whose radii are 1 and 3. Find, with proof, the
locus of all points
for which there exists points on
and
on
such that
is the midpoint of the line
segment
.
A3 Suppose that each of 20 students has made a choice of
anywhere from 0 to 6 courses from a total of 6 courses
offered. Prove or disprove: there are 5 students and 2
courses such that all 5 have chosen both courses or all 5
have chosen neither course.
A4 Let be the set of ordered triples
of distinct
elements of a finite set
. Suppose that
1.
if and only if
;
2.
if and only if
;
3.
and
are both in if and only if
and
are both in
.
Prove that there exists a one-to-one function from
to
such that implies .
Note:
is the set of real numbers.
A5 If is a prime number greater than 3 and
,
prove that the sum
of binomial coefficients is divisible by
.
A6 Let
be a constant. Give a complete descrip-
tion, with proof, of the set of all continuous functions
such that
for all
.
Note that
denotes the set of real numbers.
B1 Define a selfishset to be a set which has its own cardi-
nality (number of elements) as an element. Find, with
proof, the number of subsets of
which are
minimalselfish sets, that is, selfish sets none of whose
proper subsets is selfish.
B2 Show that for every positive integer
,
B3 Given that
, find,
with proof, the largest possible value, as a function of
(with ), of
B4 For any square matrix
, we can define
by the
usual power series:
Prove or disprove: there exists a
matrix
with
real entries such that
B5 Given a finite string of symbols and , we write
for the number of s in minus the number
of s. For example,
. We
call a string balanced if every substring of (con-
secutive symbols of) has
. Thus,
is not balanced, since it contains the sub-
string
. Find, with proof, the number of bal-
anced strings of length .
B6 Let
be the vertices of a
convex polygon which contains the origin in its inte-
rior. Prove that there exist positive real numbers and
such that
-
7/26/2019 Putnam All Questions
16/109
Solutions to the Fifty-Eighth William Lowell Putnam Mathematical Competition
Saturday, December 7, 1996
Manjul Bhargava and Kiran Kedlaya
A-1 If and are the sides of two squares with combined
area 1, then
. Suppose without loss of gen-
erality that
. Then the shorter side of a rectangle
containing both squares without overlap must be at least , and the longer side must be at least
. Hence the
desired value of is the maximum of
.
To find this maximum, we let
with . Then we are to maximize
with equality for . Hence this value is the de-
sired value of .
A-2 Let
and
be the centers of
and
, respec-
tively. (We are assuming
has radius 1 and
has
radius 3.) Then the desired locus is an annulus centered
at the midpoint of
, with inner radius 1 and outer
radius 2.
For a fixed point on
, the locus of the midpoints of
the segments for lying on is the image of
under a homothety centered at of radius
, which
is a circle of radius
. As varies, the center of this
smaller circle traces out a circle
of radius
(again
by homothety). By considering the two positions of
on the line of centers of the circles, one sees that
is
centered at the midpoint of
, and the locus is now
clearly the specified annulus.
A-3 The claim is false. There are
ways to choose
3 of the 6 courses; have each student choose a different
set of 3 courses. Then each pair of courses is chosen by
4 students (corresponding to the four ways to complete
this pair to a set of 3 courses) and is not chosen by 4
students (corresponding to the 3-element subsets of the
remaining 4 courses).
Note: Assuming that no two students choose the same
courses, the above counterexample is unique (up to per-
muting students). This may be seen as follows: Given a
group of students, suppose that for any pair of courses
(among the six) there are at most 4 students taking both,
and at most 4 taking neither. Then there are at most
pairs
, where is a student, and
is a set of two courses of which is taking either both
or none. On the other hand, if a student is taking
courses, then he/she occurs in
such
pairs
. As
is minimized for
, it follows
that every student occurs in at least
such
pairs
. Hence there can be at most
stu-
dents, with equality only if each student takes 3 courses,
and for each set of two courses, there are exactly 4 stu-
dents who take both and exactly 4 who take neither.
Since there are only 4 ways to complete a given pair
of courses to a set of 3, and only 4 ways to choose 3
courses not containing the given pair, the only way for
there to be 20 students (under our hypotheses) is if allsets of 3 courses are in fact taken. This is the desired
conclusion.
However, Robin Chapman has pointed out that the so-
lution is not unique in the problem as stated, because a
given selection of courses may be made by more than
one student. One alternate solution is to identify the 6
courses with pairs of antipodal vertices of an icosahe-
dron, and have each student pick a different face and
choose the three vertices touching that face. In this ex-
ample, each of 10 selections is made by a pair of stu-
dents.
A-4 In fact, we will show that such a function
exists withthe property that
if and only if
for some cyclic permutation
of
. We proceed by induction on the number of el-
ements in . If and , then
choose with , otherwise choose
with .
Now let be an element of
and
.
Let be the elements of labeled such that
. We claim that there ex-
ists a unique such that ,
where hereafter
.
We show existence first. Suppose no such exists; then
for all
, we have
.
This holds by property 1 for and by induction on
in general, noting that
Applying this when
, we get
,
-
7/26/2019 Putnam All Questions
17/109
contradicting the fact that
. Hence ex-
istence follows.
Now we show uniqueness. Suppose
; then for any , we have
by the assumption on
. Therefore
so
. The case
is ruled out by
and the case is similar.
Finally, we put in if , and
otherwise; an analysis similar to that
above shows that has the desired property.
A-5 (due to Lenny Ng) For
, divides
and
where the congruence
means that
is a rational number whose numerator, in reduced form,
is divisible by . Hence it suffices to show that
We distinguish two cases based on . First
suppose
, so that
. Then
since
.
Now suppose
, so that
. A similar
argument gives
A-6 We first consider the case
; we shall show in
this case must be constant. The relation
proves that is an even function. Let
be the
roots of , both of which are real. If
,
define
and
for each positiveinteger . By induction on ,
for all , so the sequence tends to a limit
which is a
root of not less than
. Of course this means
. Since for all and
,
we conclude
, so is constant on
.
If
and is defined as before, then by in-
duction,
. Note that the sequence can
be defined because
; the latter follows by noting
that the polynomial is positive at and
has its minimum at
, so both roots are greater
than . In any case, we deduce that is also constant
on
.
Finally, suppose
. Now define
. Given that
, we have
. Thus
if we had
for all , by the same argument as in
the first case we deduce
and so
.
Actually, this doesnt happen; eventually we have
, in which case by what we
have already shown. We conclude that is a constant
function. (Thanks to Marshall Buck for catching an in-
accuracy in a previous version of this solution.)
Now suppose
. Then the sequence defined
by and
is strictly increasing
and has no limit point. Thus if we define on
as any continuous function with equal values on the
endpoints, and extend the definition from
to
by the relation
, and
extend the definition further to
by the relation
, the resulting function has the desired
property. Moreover, any function with that property
clearly has this form.
B-1 Let denote the set
, and let denote
the number of minimal selfish subsets of . Then the
number of minimal selfish subsets of not containing
2
-
7/26/2019 Putnam All Questions
18/109
is equal to
. On the other hand, for any mini-
mal selfish subset of containing , by subtracting 1
from each element, and then taking away the element from the set, we obtain a minimal selfish subset
of
(since and cannot both occur in a selfish
set). Conversely, any minimal selfish subset of
gives rise to a minimal selfish subset of containing by the inverse procedure. Hence the number of min-
imal selfish subsets of
containing
is
. Thuswe obtain
. Since
, we
have , where denotes the th term of the
Fibonacci sequence.
B-2 By estimating the area under the graph of using up-
per and lower rectangles of width 2, we get
Since
, we have, upon expo-
nentiating and taking square roots,
using the fact that
.
B-3 View
as an arrangement of the numbers
on a circle. We prove that the optimal ar-
rangement is
To show this, note that if is a pair of adjacent num-
bers and is another pair (read in the same order
around the circle) with and , then the seg-
ment from to can be reversed, increasing the sum
by
Now relabel the numbers so they appear in order as fol-lows:
where without loss of generality we assume
. By considering the pairs
and
and using the trivial fact
, we
deduce
. We then compare the pairs
and
, and using that
, we deduce
. Continuing in this
fashion, we prove that
and
so
for
, i.e. that the optimal ar-
rangement is as claimed. In particular, the maximum
value of the sum is
Alternate solution: We prove by induction that the value
given above is an upper bound; it is clearly a lower
bound because of the arrangement given above. As-
sume this is the case for . The optimal arrangement
for is obtained from some arrangement for by
inserting between some pair of adjacent terms.
This operation increases the sum by
, which is an increasing function of
both and . In particular, this difference is maximal
when and equal and
. Fortunately, this
yields precisely the difference between the claimed up-
per bound for and the assumed upper bound for ,
completing the induction.
B-4 Suppose such a matrix exists. If the eigenvalues of
(over the complex numbers) are distinct, then there
exists a complex matrix such that
is
diagonal. Consequently,
is diagonal. But then
must be diagonalizable, a con-
tradiction. Hence the eigenvalues of are the same,
and has a conjugate
over the complex
numbers of the form
A direct computation shows that
Since
and
are conjugate, their eigenvalues
must be the same, and so we must have . This
implies , so that is the identity matrix,
as must be , a contradiction. Thus cannot exist.
Alternate solution (due to Craig Helfgott and AlexPopa): Define both
and
by the usual power
series. Since commutes with itself, the power series
identity
holds. But if is the given matrix, then by the above
identity,
must equal
which is
3
-
7/26/2019 Putnam All Questions
19/109
a nilpotent matrix. Thus is also nilpotent. How-
ever, the square of any
nilpotent matrix must be
zero (e.g., by the Cayley-Hamilton theorem). This is a
contradiction.
B-5 Consider a
checkerboard, in which we write an -letter string, one letter per square. If the string is
balanced, we can cover each pair of adjacent squares
containing the same letter with a
domino, and
these will not overlap (because no three in a row can
be the same). Moreover, any domino is separated from
the next by an even number of squares, since they must
cover opposite letters, and the sequence must alternate
in between.
Conversely, any arrangement of dominoes where ad-
jacent dominoes are separated by an even number of
squares corresponds to a unique balanced string, once
we choose whether the string starts with or . In
other words, the number of balanced strings is twice
the number of acceptable domino arrangements.
We count these arrangements by numbering the squares and distinguishing whether the dominoes
start on even or odd numbers. Once this is decided, one
simply chooses whether or not to put a domino in each
eligible position. Thus we have
arrangements in
the first case and
in the second, but note that
the case of no dominoes has been counted twice. Hence
the number of balanced strings is
B-6 We will prove the claim assuming only that the convex
hull of the points contains the origin in its in-
terior. (Thanks to Marshall Buck for pointing out that
the last three words are necessary in the previous sen-
tence!) Let
so that the left-hand
side of the given equation is
(1)
Now note that (1) is the gradient of the function
and so it suffices to show has a critical point. We will
in fact show has a global minimum.
Clearly we have
Note that this maximum is positive for :
if we had for all , then the subset
of the -plane would be a half-plane
containing all of the points , whose convex hull
would then not contain the origin, a contradiction.
The function
is clearly continuous on
the unit circle
, which is compact. Hence it
has a global minimum , and so for all ,
In particular,
on the disk of radius
. Since
, the infimum of
is the same over the entire -plane as over this disk,
which again is compact. Hence attains its infimal
value at some point in the disk, which is the desired
global minimum.
Noam Elkies has suggested an alternate solution as fol-
lows: for , draw the loop traced by (1) as
travels counterclockwise around the circle .
For , this of course has winding number 0 about
any point, but for large, one can show this loop has
winding number 1 about the origin, so somewhere in
between the loop must pass through the origin. (Prov-
ing this latter fact is a little tricky.)
4
-
7/26/2019 Putnam All Questions
20/109
The Fifty-Eighth William Lowell Putnam Mathematical Competition
Saturday, December 6, 1997
A1 A rectangle,
, has sides
and
. A triangle
has
as the intersection of the al-
titudes,
the center of the circumscribed circle,
the
midpoint of
, and
the foot of the altitude from .
What is the length of
?
A2 Players
are seated around a table, and
each has a single penny. Player 1 passes a penny to
player 2, who then passes two pennies to player 3.
Player 3 then passes one penny to Player 4, who passes
two pennies to Player 5, and so on, players alternately
passing one penny or two to the next player who still
has some pennies. A player who runs out of pennies
drops out of the game and leaves the table. Find an in-
finite set of numbers
for which some player ends upwith all
pennies.
A3 Evaluate
A4 Let be a group with identity and a
function such that
whenever
. Prove that thereexists an element such that
is a homomorphism (i.e.
for all
).
A5 Let denote the number of ordered
-tuples of posi-
tive integers
such that
. Determine whether
is even or odd.
A6 For a positive integer
and any real number , define
recursively by
,
, and for
,
Fix
and then take to be the largest value for which
. Find
in terms of
and ,
.
B1 Let
denote the distance between the real number
and the nearest integer. For each positive integer
,
evaluate
(Here
denotes the minimum of and
.)
B2 Let be a twice-differentiable real-valued function sat-
isfying
where
for all real
. Prove that
is
bounded.
B3 For each positive integer
, write the sum
in the form
, where
and
are relatively prime
positive integers. Determine all
such that 5 does not
divide
.
B4 Let
denote the coefficient of
in the expansion
of
. Prove that for all [integers]
,
B5 Prove that for
,
terms
terms
B6 The dissection of the 345 triangle shown below (into
four congruent right triangles similar to the original) has
diameter
. Find the least diameter of a dissection of
this triangle into four parts. (The diameter of a dissec-
tion is the least upper bound of the distances betweenpairs of points belonging to the same part.)
-
7/26/2019 Putnam All Questions
21/109
Solutions to the Fifty-Eighth William Lowell Putnam Mathematical Competition
Saturday, December 6, 1997
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng
A1 The centroid
of the triangle is collinear with and
(Euler line), and the centroid lies two-thirds of the
way from to
. Therefore
is also two-thirds of
the way from to
, so
. Since the triangles
and
are similar (theyre right triangles and
), we have
, or
. Now
, but
, so
A2 We show more precisely that the game terminates with
one player holding all of the pennies if and only if
or
for some
. First sup-
pose we are in the following situation for some
.
(Note: for us, a move consists of two turns, starting
with a one-penny pass.)
Except for the player to move, each player has
pennies;
The player to move has at least pennies.
We claim then that the game terminates if and only if
the number of players is a power of 2. First supposethe number of players is even; then after complete
rounds, every other player, starting with the player who
moved first, will have more pennies than initially, and
the others will all have 0. Thus we are reduced to the
situation with half as many players; by this process, we
eventually reduce to the case where the number of play-
ers is odd. However, if there is more than one player,
after two complete rounds everyone has as many pen-
nies as they did before (here we need
), so the
game fails to terminate. This verifies the claim.
Returning to the original game, note that after one com-
plete round,
players remain, each with 2 pennies
except for the player to move, who has either 3 or 4 pen-nies. Thus by the above argument, the game terminates
if and only if
is a power of 2, that is, if and only
if
or
for some .
A3 Note that the series on the left is simply
.
By integration by parts,
and so by induction,
Thus the desired integral is simply
A4 In order to have for all , we must in par-
ticular have this for , and so we take
.
We first note that
and so
commutes with
for all
. Next, we
note that
and using the commutativity of , we deduce
or
, as desired.
A5 We may discard any solutions for which
, since
those come in pairs; so assume
. Similarly, wemay assume that
,
,
,
.
Thus we get the equation
Again, we may assume
and
, so we get
; and
, so
. This implies that
, which by
counting has 5 solutions. Thus
is odd.
A6 Clearly
is a polynomial in of degree , so it suf-
fices to identify values of for which
. We
claim these are
for
; in
this case,
is the coefficient of
in the polynomial
. This can be verified bynoticing that
satisfies the differential equation
(by logarithmic differentiation) or equivalently,
-
7/26/2019 Putnam All Questions
22/109
and then taking the coefficient of
on both sides:
In particular, the largest such is , and
for
.
Greg Kuperberg has suggested an alternate approach to
show directly that
is the largest root, withoutcomputing the others. Note that the condition
states that
is an eigenvector of the matrix
otherwise
with eigenvalue . By the Perron-Frobenius theorem,
has a unique eigenvector with positive entries, whose
eigenvalue has modulus greater than or equal to that of
any other eigenvalue, which proves the claim.
B1 It is trivial to check that
for
, that
for
, that
for
, and that
for
. Therefore the desired sum is
B2 It suffices to show that
is bounded for ,
since
satisfies the same equation as
. But
then
so that
for
.
B3 The only such are the numbers 14, 2024, 100104,
and 120124. For the proof let
and introduce the auxiliary function
It is immediate (e.g., by induction) that
(mod ) for
(mod 5) re-
spectively, and moreover, we have the equality
where denotes the largest integer such that
. We wish to determine those such that the
above sum has nonnegative 5valuation. (By the 5
valuation of a number we mean the largest integer
such that
is an integer.)
If
, then the last term in the above sum
has 5valuation , since
,
,
each have valu-
ation 0; on the other hand, all other terms must have
5valuation strictly larger than . It follows that
has 5valuation exactly ; in particular,
has non-
negative 5valuation in this case if and only if ,
i.e., , 2, or 3.
Suppose now that
. Then we must also
have
. The former condition im-
plies that the last term of the above sum is
, which has 5valuation
.
It is clear that
(mod 25); hence if
equals 20 or 24, then the secondtolast term
of the above sum (if it exists) has valuation at least
. The thirdtolast term (if it exists) is of
the form
, so that the sum of the last term andthe third to last term takes the form
.
Since
can be congruent only to 0,1, or -1 (mod 5),
and
(mod 5), we conclude that the sum of the
last term and thirdtolast term has valuation
,
while all other terms have valuation strictly higher.
Hence
has nonnegative5valuation in this case only
when
, leading to the values
(arising from
), 20,24 (arising from
and
and 24 resp.), 101, 102, 103, and 104 (arising from
,
) and 120, 121, 122, 123, and
124 (arising from ,
).
Finally, suppose
and
, 22,
or 23. Then as before, the first condition implies thatthe last term of the sum in (*) has valuation
,
while the second condition implies that the secondto
last term in the same sum has valuation
. Hence
all terms in the sum (*) have 5valuation strictly higher
than
, except for the secondtolast term, and
therefore
has 5valuation
in this case. In
particular,
is integral (mod 5) in this case if and only
if
, which gives the additional values
, 22,
and 23.
B4 Let
be the given sum (note that
is nonzero precisely for
. Since
we have
2
-
7/26/2019 Putnam All Questions
23/109
By computing
, we may eas-
ily verify by induction that
and
for all
. (Alternate so-
lution suggested by John Rickert: write
, and note note that is the
coefficient of
in
.)
B5 Define the sequence
,
for .
It suffices to show that for every
,
for some
. We do this by induction on , with
being obvious.
Write
, where is odd. It suffices to show
that
modulo
and modulo , for some
. For the former, we only need
,
but clearly
by induction on . For the lat-
ter, note that
as long as
, where is the Eu-
ler totient function. By hypothesis, this occurs for some
. (Thanks to Anoop Kulkarni for
catching a lethal typo in an earlier version.)
B6 The answer is
. Place the triangle on the carte-
sian plane so that its vertices are at
. It is easy to check that the five points
and
are
all in the triangle and have distance at least
apart
from each other (note that
); thus any dis-
section of the triangle into four parts must have diame-
ter at least
.
We now exhibit a dissection with least diameter
.(Some variations of this dissection are possible.) Put
,
,
,
, and divide
into
the convex polygonal regions
,
,
,
; each region has diameter
, as can be
verified by checking the distance between each pair of
vertices of each polygon. (One need only check for the
pentagon: note that and
are contained in
circular sectors centered at and
, respectively, of ra-
dius
and angle less than
, and that
is
a rectangle with diagonal
.)
3
-
7/26/2019 Putnam All Questions
24/109
The 59th William Lowell Putnam Mathematical Competition
Saturday, December 5, 1998
A1 A right circular cone has base of radius 1 and height 3.A cube is inscribed in the cone so that one face of the
cube is contained in the base of the cone. What is the
side-length of the cube?
A2 Lets be any arc of the unit circle lying entirely in thefirst quadrant. LetA be the area of the region lyingbelows and above thex-axis and let B be the area ofthe region lying to the right of they-axis and to the leftofs. Prove thatA +B depends only on the arc length,and not on the position, ofs.
A3 Let fbe a real function on the real line with continuousthird derivative. Prove that there exists a pointa suchthat
f(a)
f(a)
f(a)
f(a)
0.
A4 Let A1 = 0 and A2 = 1. For n > 2, the num-berAn is defined by concatenating the decimal expan-sions ofAn1 and An2 from left to right. For ex-ample A3 = A2A1 = 10, A4 = A3A2 = 101,A5 = A4A3 = 10110, and so forth. Determine allnsuch that11 divides An.
A5 LetFbe a finite collection of open discs in R2 whoseunion contains a set E R2. Show that there is apairwise disjoint subcollection D1, . . . , Dn inF suchthat
E nj=13Dj.Here, ifD is the disc of radius rand centerP, then3Dis the disc of radius3rand centerP.
A6 LetA, B,Cdenote distinct points with integer coordi-nates in R2. Prove that if
(|AB| + |BC|)2 0.
B2 Given a point(a, b)with0< b < a, determine the minimum perimeter of a triangle with one vertex at (a, b)one on thex-axis, and one on the liney =x. You maassume that a triangle of minimum perimeter exists.
B3 let H be the unit hemisphere{(x,y,z) : x2 +y2 +z2 = 1, z 0}, C the unit circle{(x,y, 0) : x2 +y2 = 1}, and Pthe regular pentagon inscribed in CDetermine the surface area of that portion ofH lyin
over the planar region insideP, and write your answein the form A sin + B cos , where A,B,,are reanumbers.
B4 Find necessary and sufficient conditions on positive in
tegersmandnso that
mn1
i=0
(1)i/m+i/n = 0.
B5 LetN
be the positive integer with 1998 decimal digits
all of them 1; that is,
N= 1111 11.
Find the thousandthdigit after the decimal point of
N
B6 Prove that, for any integersa, b, c, there exists a positive integern such that
n3 + an2 + bn + cis not a
integer.
-
7/26/2019 Putnam All Questions
25/109
Solutions to the 59th William Lowell Putnam Mathematical Competition
Saturday, December 5, 1998
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng
A1 Consider the plane containing both the axis of the cone
and two opposite vertices of the cubes bottom face.
The cross section of the cone and the cube in this plane
consists of a rectangle of sides and inscribed in
an isosceles triangle of base and height , where is
the side-length of the cube. (The side of the rect-
angle lies on the base of the triangle.) Similar triangles
yield
, or
A2 First solution: to fix notation, let be the area of re-
gion
, and be the area of
; further let
denote the area of sector
, which only depends
on the arc length of
. If
denotes the area of tri-angle
, then we have
and
. But clearly
and
, and so
.
O
D
E
F G
H
I
Second solution: We may parametrize a point in by
any of ,
, or
. Then and
are
just the integrals of
and
over the appropriate
intervals; thus
is the integral of
(mi-
nus because the limits of integration are reversed). But
, and so
is precisely the
radian measure of . (Of course, one can perfectly well
do this problem by computing the two integrals sepa-
rately. But whats the fun in that?)
A-3 If at least one of
,
,
, or
van-
ishes at some point
, then we are done. Hence wemay assume each of , , , and
is either strictly positive or strictly negative on the
real line. By replacing by if necessary,
we may assume ; by replacing by
if necessary, we may assume . (No-
tice that these substitutions do not change the sign of
.) Now
implies that
is increasing, and
implies that
is convex, so that
for all
and
. By letting
increase in the latter inequality,
we see that
must be positive for sufficiently
large ; it follows that
for all . Similarly,
and
imply that
for all
. Therefore
for all , and
we are done.
A-4 The number of digits in the decimal expansion of
is the Fibonacci number , where
,
,
and
for
. It follows that
the sequence
, modulo 11, satisfies the recursion
. (Notice that the recur-
sion for
depends only on the value of
modulo2.) Using these recursions, we find that and
modulo 11, and that
and
mod-
ulo 2. It follows that
(mod 11) for all
. We find that among
, and
, only
vanishes modulo 11. Thus 11 divides
if and only if
for some nonnegative integer
.
A5 Define the sequence by the following greedy algo-
rithm: let
be the disc of largest radius (breaking ties
arbitrarily), let
be the disc of largest radius not meet-
ing
, let
be the disc of largest radius not meeting
or
, and so on, up to some final disc
. To see
that
, consider a point in ; if it lies inone of the
, we are done. Otherwise, it lies in a disc
of radius
, which meets one of the
having radius
(this is the only reason a disc can be skipped
in our algorithm). Thus the centers lie at a distance
, and so every point at distance less than
from the center of lies at distance at most
from the center of the corresponding
.
A6 Recall the inequalities
(AM-GM) and
(Law of Sines).
Also recall that the area of a triangle with integer
coordinates is half an integer (if its vertices lie at
, the area is
), and that
if and have integer coordinates, then
is an
integer (Pythagoras). Now observe that
and that the first and second expressions are both inte-
gers. We conclude that
-
7/26/2019 Putnam All Questions
26/109
, and so
; that is, is a right angle and
,
as desired.
B1 Notice that
(difference of squares). The latter is easily seen (e.g.,
by AM-GM) to have minimum value 6 (achieved at
).
B2 Consider a triangle as described by the problem; la-
bel its vertices
so that
, lies on
the -axis, and
lies on the line
. Further
let
be the reflection of in the
-axis,
and let
be the reflection of in the line
. Then and
, and so
the perimeter of
is
. It is clear that
this lower bound can be achieved; just set (resp.
)
to be the intersection between the segment and the
-axis (resp. line
); thus the minimum perimeter
is in fact
.
B3 We use the well-known result that the surface area of
the sphere cap
is simply
. (This result is easily verified using
calculus; we omit the derivation here.) Now the desired
surface area is just minus the surface areas of five
identical halves of sphere caps; these caps, up to isome-
try, correspond to
being the distance from the center
of the pentagon to any of its sides, i.e.,
.
Thus the desired area is
(i.e.,
).
B4 For convenience, define
, so that
the given sum is
. If
and are both odd, then
is the sum of an
odd number of s, and thus cannot be zero. Now
consider the case where and have opposite par-
ity. Note that
for all
integers
. Thus
and
; this implies that
is odd, and so
for all . It follows
that
if and
have opposite parity.
Now suppose that
and
are both even.Then
for all , so
can be computed
as twice the sum over only even indices:
Thus
vanishes if and only if
vanishes
(if
, then and have opposite parity
and so
also vanishes).
Piecing our various cases together, we easily deduce
that
if and only if the highest powers of 2
dividing and
are different.
B5 Write
. Then
where
. Now the digits after the deci-
mal point of
are given by
, while the
digits after the decimal point of
are given by
. It follows that the first 1000
digits of are given by
; in particu-
lar, the thousandth digit is .
B6 First solution: Write
. Note
that
and
have the same parity, and recall
that any perfect square is congruent to 0 or 1 (mod 4).
Thus if and
are perfect squares, they are
congruent mod 4. But
(mod4), which is not divisible by 4 if
and
have opposite
parity.
Second solution: We prove more generally that for any
polynomial
with integer coefficients which is not
a perfect square, there exists a positive integer such
that
is not a perfect square. Of course it suffices
to assume
has no repeated factors, which is to say
and its derivative
are relatively prime.
In particular, if we carry out the Euclidean algorithm
on
and
without dividing, we get an in-
teger (the discriminant of
) such that the great-
est common divisor of
and
divides
for any . Now there exist infinitely many primes
such that divides for some : if there were
only finitely many, say,
, then for any di-
visible by
, we have
, that is,
is not divisible
by
, so must be
, but then
takes some
value infinitely many times, contradiction. In particu-
lar, we can choose some such not dividing
, and
choose such that
divides
. Then
(write out the Taylor series
of the left side); in particular, since does not divide
, we can find some such that
is di-
visible by but not by
, and so is not a perfect square.
Third solution: (from David Rusin, David Savitt, and
Richard Stanley independently) Assume that
is a square for all
. For sufficiently large
,
2
-
7/26/2019 Putnam All Questions
27/109
thus if is a large even perfect square, we have
. We conclude this is an
equality of polynomials, but the right-hand side is not a
perfect square for an even non-square, contradiction.
(The reader might try generalizing this approach to ar-
bitrary polynomials. A related argument, due to Greg
Kuperberg: write
as
times a
power series in and take two finite differences to
get an expression which tends to 0 as , contra-
diction.)
Note: in case
has no repeated fac-
tors, it is a square for only finitely many , by a theorem
of Siegel; work of Baker gives an explicit (but large)
bound on such . (I dont know whether the graders
will accept this as a solution, though.)
3
-
7/26/2019 Putnam All Questions
28/109
The 60th William Lowell Putnam Mathematical Competition
Saturday, December 4, 1999
A-1 Find polynomials
,
, and
, if they exist,such that for all
,
if
if
if
.
A-2 Let
be a polynomial that is nonnegative for all
real
. Prove that for some , there are polynomials
) such that
A-3 Consider the power series expansion
Prove that, for each integer
, there is an integer
such that
A-4 Sum the series
A-5 Prove that there is a constant such that, if
is a
polynomial of degree 1999, then
A-6 The sequence
is defined by
and, for
,
Show that, for all n,
is an integer multiple of .
B-1 Right triangle has right angle at and
; the point is chosen on
so that
; the point is chosen on
so that
.
The perpendicular to
at meets
at
. Evalu-
ate
.
B-2 Let
be a polynomial of degree such that
, where
is a quadratic polynomial and
is the second derivative of
. Show that if
has at least two distinct roots then it must have
distinct roots.
B-3 Let
. For
, let
where the sum ranges over all pairs
of positive
integers satisfying the indicated inequalities. Evaluate
B-4 Let
be a real function with a continuous third deriva-
tive such that
are positive forall
. Suppose that
for all
. Show that
for all
.
B-5 For an integer
, let
. Evaluate the
determinant of the
matrix
, where
is
the
identity matrix and
has entries
for all
.
B-6 Let
be a finite set of integers, each greater than 1.
Suppose that for each integer there is some
such that
or
. Show that
there exist
such that
is prime.
-
7/26/2019 Putnam All Questions
29/109
Solutions to the 60th William Lowell Putnam Mathematical Competition
Saturday, December 4, 1999
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng
A1 Note that ifr(x)and s(x)are any two functions, then
max(r, s) = (r+s+ |r s|)/2.
Therefore, ifF(x)is the given function, we have
F(x) = max{3x 3, 0} max{5x, 0} + 3x+ 2= (3x 3 + |3x+ 3|)/2
(5x+ |5x|)/2 + 3x+ 2= |(3x+ 3)/2| |5x/2| x+1
2,
so we may setf(x) = (3x+ 3)/2,g(x) = 5x/2, andh(x) = x+ 12 .
A2 First solution: First factorp(x) = q(x)r(x), whereqhas all real roots andr has all complex roots. Noticethat each root ofq has even multiplicity, otherwise pwould have a sign change at that root. Thusq(x)has asquare roots(x).
Now write r(x) =k
j=1(x aj)(x aj) (possiblebecauser has roots in complex conjugate pairs). Writek
j=1(x aj) = t(x) + iu(x) with t, x having realcoefficients. Then forxreal,
p(x) = q(x)r(x)
=s(x)
2
(t(x) +iu(x))(t(x) +iu(x))= (s(x)t(x))2 + (s(x)u(x))2.
(Alternatively, one can factor r(x) as a product ofquadratic polynomials with real coefficients, write each
as a sum of squares, then multiply together to get a sum
of many squares.)
Second solution: We proceed by induction on the de-
gree ofp, with base case where p has degree 0. As inthe first solution, we may reduce to a smaller degree
in case p has any real roots, so assume it has none.Then p(x) > 0 for all real x, and since p(x) forx
,phas a minimum valuec. Nowp(x)
c
has real roots, so as above, we deduce that p(x) cisa sum of squares. Now add one more square, namely
(
c)2, to getp(x)as a sum of squares.
A3 First solution: Computing the coefficient ofxn+1 in theidentity(12x x2)m=0amxm = 1 yields therecurrencean+1 = 2an + an1; the sequence{an}is then characterized by this recurrence and the initial
conditionsa0 = 1, a1= 2.
Define the sequence {bn} by b2n = a2n1 +
a2n, b2n+1 = an(an1+an+1).Then
2b2n+1+b2n= 2anan+1+ 2an1an+a2n1+a
2n
= 2anan+1+an1an+1+a2n
=a2n+1+a2n = b2n+2,
and similarly 2b2n + b2n1 = b2n+1, so that{bn}satisfies the same recurrence as{an}. Since furtherb0 = 1, b1 = 2 (where we use the recurrence for {an}to calcul