01 Number System Aptitude Trick-Book AptiDude

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CHAPTER-1 NUMBER SYSTEM




NUMBER SYSTEMAs they say, before you learn elementary algebrayou learn arithmetic. The subject matter ofarithmetic consists largely of various "numbersystems”. Let’s hover over some very interestingthings on number systems.Interesting Basics1. Any terminating or recurring decimal is arational number.2. The product of ‘n’ consecutive natural numbersis always divisible by n!3. Consecutive integers can be written as n, n+1,n+2, n+3 ……. and so on. Similarly, consecutiveodd and even integers are written as n, n+2,n+4………..and so on.4. How many times a digit ‘n’ appears in a certainrange (except 0)?a) From 1 – 9, the digit ‘n’ is used 1 timeb) From 1 – 99, the digit ‘n’ is used 20 times.c) From 1 – 999, the digit ‘n’ is used 300 timesd) From 1 – 9999, the digit ‘n’ is used 4000times and so on…

Check the pattern! 1, 20, 300, 4000, 50000….Now, if someone asks, how many 8s (or anydigit) are present in 1 to 999? Just say 300.Cheers!5. Number of digits between two numbers:a) Number of single digit natural numbers(i.e. from 1 – 9) = 9b) Number of two digits natural numbers(i.e. from 10 – 99) = 90c) Number of three digits natural number(i.e. from 100 – 999) = 900

d) Number of 4 digits Natural Number (i.e.from 1000 to 9999) = 9000 and so on…Check the pattern 9, 90, 900, 9000, 90000….

6. The product of ‘n’ digits and ‘k’ digits numbercan be a maximum of (n+k) digit number. Forexample, if a 2 digit number is multiplied by asingle digit number the product can be ofmaximum 3 digits (2+1)

ApplicationIf you write first 252 natural numbers in astraight line, how many times do you write thedigit 4?Solution: In the first 99 natural numbers, digit 4comes 20 times. Similarly, from 100 to 199, digit4 comes 20 times. Now from 200 to 299, digit 4comes again 20 times out of which we need tosubtract 5 numbers (254, 264, 274, 284 and294).Therefore, total number of times that we writethe digit 4 = 20 + 20 + 20 − 5 = 55

A CAT admit card has a 6-digit number. If wemultiply the 1st two digits with 3, we get all 1’s.If we multiply the next two digits with 6, we getall 2’s. If we multiply the last two digits with 9,we get all 3’s. Find the number on the admitcard.Solution:1113 = 37 , 2226 = 37, 3339 = 37Thus the number is 373737

Application




7. The product of 3 consecutive natural numbersis divisible by 6. If the 1st number is an evennumber, then the product is divisible by 24also.

Square Number Check1. Square of any natural number can be written inthe form of 3n or 3n+1 or 4n or 4n+1.2. Any square number never ends [unit digit]with 2, 3, 7 or 8.3. Ten’s digit of any square number is alwayseven [except when unit digit of that squarenumber is ‘6’]. For example. 92 = 81, 192 = 361,etc except 142 = 196, 162 = 256, etc4. The digital sum of any square number willalways be 1 or 4 or 7 or 9. For example;a) 142 = 196. Digital Sum of 196 = 1 + 9 + 6 =16 = 1 + 6 = 7

b) 882 = 7744. Digital Sum of 7744 = 7 + 7 + 4+ 4= 22 = 2 + 2 = 4

Squaring Techniques

1. Number close to 50 but less than 50

Step 1: Quickly determine how much is the'difference' of the given number from 50Step 2: Subtract that 'difference' from the 'keynumber' 25Step 3: Simply put the square of that 'difference' infront of the resultant you obtained in step 2

492 The difference from 50 is 1;=> 25 - 1 = 24; => 012 = 01∴ 492 = 2401382(TheCarry-Case) The difference from 50 is 12;=> 25 - 12 = 13; => 122 = 144∴ 492 = 13144 = (13+1)44 = 1444

2. Number close to 50 but more than 50

Step 1: Quickly determine how much is the'difference' of the given number from 50Step 2: Add that 'difference' to the 'key number' 25Step 3: Simply put the square of that 'difference' infront of the resultant you obtained in step 2

512 The difference from 50 is 1;=> 25 + 1 = 26; => 012 = 01∴ 492 = 2601622(TheCarry-Case)The difference from 50 is 12;=> 25 + 12 = 37; => 122 = 144∴ 492 = 37144 = (37+1)44 = 3844

Check out this interesting pattern: The number of 1sin the product is equal to the number of digits in themultiplicands. 3 x 37 = 111 33 x 3367 = 111,111 333 x 333667 = 111,111,111 3333 x 33336667= 111,111,111,111 33333 x 3333366667= 111,111,111,111,111 333333 x 333333666667 = 111,111,111,111,111,111__________________________________________________________________When 6 is multiplied with 7 it always gives 4 and 2. 6 x 7 = 42 66 x 67 = 4422 666 x 667 = 444222 6666 x 6667 = 44442222

Numbers Magic




3. Number close to 100 but less than 100

Step 1: Quickly determine how much is the‘difference’ of the given number from 100Step 2: Subtract that 'difference' from the 'givennumber itself'Step 3: Simply put the square of that 'difference' infront of the resultant you obtained in step 2

992 The difference from 100 is 1;=> 99 - 1 = 98; => 012 = 01∴ 992 = 9801882(TheCarry-Case) The difference from 100 is 12;=> 100 - 12 = 88; => 122 = 144∴ 882 = 76144 = (76+1)44 = 7744

4. Number close to 100 but more than 100

Step 1: Quickly determine how much is the‘difference’ of the given number from 100Step 2: Add that 'difference' to the 'given numberitself'Step 3: Simply put the square of that 'difference' infront of the resultant you obtained in step 2

1012 The difference from 100 is 1;=> 101 + 1 = 102; => 012 = 01∴ 1012 = 102011122(TheCarry-Case)The difference from 100 is 12;=> 112 + 12 = 124; => 122 = 144∴ 1122 = 124144= (124+1)44 = 12544

Note: This case of 100 is applicable to thenumbers close to 1000, 10000, 100000 and soon.

5. Number with unit digit as ‘5’05² = 02515² = 22525² = 62535² = 1225Observed any pattern?Simple! It’s obvious that, square of the numberswith unit digit 5 will end with 25. And the restdigits in the resultant will be the product of thatother digits and its next digit in the given number.35² -> (3 x 4) | 25 = 122545² -> (4 x 5) | 25 = 2025115² -> (11 x 12) | 25 = 13225

6. Finding squares with 3-digits numbersSuppose you have to find square of 192.Step 1: Just round up/down the given number tonearest multiple of 100.Step 2: Multiply the two numbers, which iscomparatively easy then before.Step 3: Simply add the square of the ‘differencenumber’Take example: 1922+8 200192 36800 + 82 =36800 + 64= 36864-8 184Take another example: 8122-12 800812 659200 +122 = 659344+12 824




Approximating Square Roots

Unknown Square root

= Known Sq Root – Known Sq – Unknown Sq2 * Known Sq Root

(The known square selected should be as near tothe given unknown number as possible. Moreclosely the selected known square number is,more correct the approximation is)For Example: Find √Solution:64 is the closest perfect square known near to 58Just put the values in the above mentioned formula√ = √ – 64 – 58

2 * √√ = – 6

2 *√ = – 3

8√ = 7.625

Finding in Calculator we get √ = 7.615

So, our approximation was very much near tothe answer

You can try it for 2-digit, 3-digit, 4 digit……. roots also

Cubing TechniquesSuppose you have to find cube of 24Step 1: Find the ratio of two given numbers i.e. 2:4or 1:2Step 2: write the cube of first digit of the numberi.e. 2 * 2 * 2 = 8Step 3: write numbers in a row of 4 terms in such away that the first one is the cube of the first digitand remaining three are obtained in a geometricprogression with common ratio as the ratio of theoriginal two digits i.e. the row will be like this8 16 32 64Step 4: Write twice the values of 2nd and 3rdterms under the terms respectively in second row8 16 32 6432 64Step 5: Add the numbers column wise (Right toLeft) and follow carry over process till the end asfollows

8 16 32 6432 6413 8 2 4

So Cube of 24 = 13824

(This method is applicable for finding cubes of2-Digit Numbers only)

Carryover 6

32+64+6=102Carryover 10

16+32+10=58Carryover 5




Finding the Cube root of a Perfect CubeNumberMemorize the first ten cubes, in following mannerNumber Cube Unit’s Digit

1 1 12 8 83 27 74 64 45 125 56 216 67 343 38 512 29 729 910 1000 0Now, Ask your friend to secretly pick any two-digitnumber and then have him or her use a calculatorto cube it. Let's say he picks 76. So using thecalculator he computes 76 x 76 x 76Then, He asks you to tell him the cube root of 438,976.You can revert him the answer in 5 seconds:Number Last 3 digits First 3 digits438, 976 976 438Now look in the table above and check which cubeis less than the first 3 digits of the given number,its 343, and 343 is the cube of 7………so your ten’sdigit of the cube root is 7Now look at the unit’s place of last 3 digits of thegiven number, its 6………..check from the abovetable which number from 1-10 gives you the unit’sdigit 6 in cube……………it is 63 = 216, hence theunit’s digit of the cube root is 6So, you can tell your friend that √438,9763 = 76

Prime Numbers - Interesting Basics1. 1 is not a prime, 2 is the smallest and the onlyeven prime number. 3, 5, 7 are the only 3consecutive odd prime numbers.2. Every prime number greater than 3 can bewritten in the form of (6k+1) or (6k-1).However, the converse of this is not always true.For e.g. even though 25 can be expressed in theform 6k + 1 where k = 4, it is not a primenumber.Take 7 = 6 x 1 + 1; Take 47 = 6 x 8 – 13. The remainder of the division of the square of aprime number p ≥ 5 divided by 12 or 24 is 1.4. There are 25 prime numbers in between 1 to100. Likewise;Number Range Number of Primes101-200 21201-300 16301-400 16401-500 17501-600 14601-700 16701-800 14801-900 15901-1000 145. All prime numbers except 2 and 5 ends in 1, 3, 7or 96. Two numbers are co-prime if there HCF is 1, forexample 21 & 25.7. The squares of prime numbers have exactlythree divisors i.e. 1, the prime number and thesquare itself.




Prime Number Check1. The number isn’t a prime number if:a) It ends with any number except 1,3,7,9b) If it is divisible by 3 (Check whether sum of thedigits is divisible by 3)2. Primality TestFor example, to check whether 179 is prime ornot?a) Step 1: If n is not a perfect square, find asuitable value of n’ which is a perfect squareclosest to n and greater than n. (Here, 179 isnot a perfect square. 196 is greater than 179and is the closest perfect square, so n’ = 196)b) Step 2: √n′= √196 =14, all prime numbersbelow 14 are 2, 3, 5, 7, 11, 13c) Step 3: If given number 179 is divisible by anyof the above prime numbers, then the givennumber is not a prime number. As 179 is notdivisible by any of the listed numbers, so 179 isa prime number.HCF/LCM of NumbersMost of us are aware about the Long DivisionMethods and Prime Factorization Methods forfinding the HCF and LCM of numbers. Here aresome methods other than those which arerelatively quick and easy.Example 1:Find the HCF and LCM of 42 and 96.SolutionWe make a fraction and simplify it as much as wecan as follows:

= =

(Be sure that no more simplification is possible)Therefore, HCF = 42 / 7 = 6LCM = 42 X 16 = 672.It is advisable to double checkHCF = 6 = 96 / 16...........YESLCM = 672 = 96 X 7............YESExample 2:Find the HCF and LCM of 120 and 36.

= = =

Therefore, HCF = 120 / 10 = 12LCM = 120 X 3 = 360Double check ! ! !HCF =12 = 36 / 3...........YESLCM = 360 = 36 X 10............YESAnother Interesting Way to Find HCF of morethan two given numbersSuppose you have been asked to find HCF of 220,280, and 375.Solution:220 280 375Find the difference between any two numbersamong the given three, i.e.Let’s say 280 – 220 = 60Then, write the factors/divisors of 60 indescending order as follows:60 = 60, 30, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1Then check which divisor exactly divides all thegiven three numbers (220, 280, and 375) in thesame descending order. The largest divisor which




will first exactly divide all the given three numberswill be the HCF of the numbers.In this case, the HCF will be 5, as 5 is the firstlargest divisor which will exactly divide all threenumbers.(This Technique is very much handy in cases wherethe difference between two numbers will be primenumber. In such cases, you will have to check thedivisibility with only the prime number so found, if itis exactly dividing the given numbers, then thatprime number will be the HCF, otherwise the HCF ofgiven Numbers will be 1)Example:Find HCF of 483, 502 and 546?Solution: Here, 502-483 = 19 (Prime Number)19 don’t divide all three given numbers.Hence, HCF = 1

APPLICATIONS of HCF / LCM of NUMBERS

Find the GREATESTNUMBER that will exactlydivide x, y, z. Required number= H.C.F. of x, y, and zFind the GREATESTNUMBER that will divide x,y and z leaving remaindersa, b and c respectively.

Required number= H.C.F. of (x – a), (y –b) and (z – c).Find LEAST NUMBERwhich is exactly divisibleby x, y and z. Required number= L.C.M. of x, y and zFind LEAST NUMBERwhich when divided by x, yand z leaves theremainders a, b and crespectively.

Then, it is alwaysobserved that(x – a) = (z – b) = (z –c) = K (say)∴ Required number= (L.C.M. of x, y, z) – K

Find LEAST NUMBERwhich when divided by x, yand z leaves the sameremainder ‘r’ each case.Required number= (L.C.M. of x, y, z) + r

Find the GREATESTNUMBER that will divide x,y and z leaving the sameremainder in each case.Required number= H.C.F of (x – y), (y –z) and (z – x).

Properties of HCF / LCM1. Product of two numbers = L.C.M. × H.C.F.2. If ratio of numbers is a : b and H is the HCF

of the numbers Then LCM of the numbers =H × a × b = HCF × Product of the ratios.3. H.C.F of fractions =H.C.F. of numerators / L.C.M. of denominators4. L.C.M. of fractions =L.C.M. of numerators / H.C.F. of denominators5. If HCF (a, b) = H1 and HCF (c, d) = H2, then HCF(a, b, c, d) = HCF (H1, H2).6. LCM is always a multiple of HCF of thenumbersExample 1: In a farewell party, some students aregiving pose for photograph, If the students stand at4 students per row, 2 students will be left if theystand 5 per row, 3 will be left and if they stand 6per row 4 will be left. If the total number ofstudents are greater than 100 and less than 150,how many students are there?Solution:If ‘N’ is the number of students, it is clear from thequestion that if N is divided by 4, 5, and 6, itproduces a remainders of 2, 3, & 4 respectively.Since (4 – 2) = (5 – 3) = (6 – 4) = 2, the leastpossible value of N is LCM (4, 5, 6) – 2 = 60 – 2, =58. But, 100 < N < 150. So, the next possible valueis 58 + 60 = 118.Example 2: There are some students in the class.Mr X brought 130 chocolates and distributed tothe students equally, then he was left with somechocolates. Mr Y brought 170 chocolates and




distributed equally to the students. He was alsoleft with the same no of chocolates as Mr X wasleft. Mr Z brought 250 chocolates, did the samething and left with the same no of chocolates. Whatis the max possible no of students that were in theclass?Solution:The question can be stated as, what is the highestnumber, which divides 130, 170 and 250 gives thesame remainder, i.e.HCF [(170 −130), (250 −170), (250 −130)] .i.e.HCF (40, 80, 120) = 40.Concept of CyclicityCyclicity of any number is about the last digit ofthe number and how they appear in a certaindefined manner.Example: What is the last digit / unit digit of271001?These type of questions can be solved by knowingthe cyclicity of numbers from 1-10.Table Showing Unit Digit of a Number fordifferent exponents:

UnitDigit(N)

N1 N2 N3 N4 N5 Cyclicity1 1 1 1 1 1 12 2 4 8 6 2 43 3 9 7 1 3 44 4 6 4 6 4 25 5 5 5 5 5 16 6 6 6 6 6 17 7 9 3 1 7 4

8 8 4 2 6 8 49 9 1 9 1 9 210 0 0 0 0 0 1So, here we see that for every number from 1-10,unit digit for their different exponents repeatsafter certain intervals and this interval is calledcyclicity. It is very helpful in finding the units digitof any number.

Divisibility RulesThe process to check whether a natural number isdivisible by a prime number ‘p’ is as follows:Step 1: Find the smallest multiple of p, which is ofthe form (10k + 1) or (10k – 1), where k is anatural number.Step 2: If it is of the form (10k – 1), then we checkwhether (x + ky) is divisible by p or not, and if it isof the form (10k + 1) then we check whether (x –ky) is divisible by p or not. Here y is the unit’s digitof the given prime number and x is the numberformed by rest of the digits of the given primenumber. The process can be repeated if the givennumber is having more than 3 digits.

TEN’s DIGIT CYCLICITYJust as the unit’s digit of a number thatis raised to varying powers repeatsitself in cycles, similarly the ten’s digitalso shows cyclicity.Digits Tens Digit Cyclicity2, 3, 8 204, 9 106 57 45 1




For a numberto be divisible

by:It should satisfy the following conditions:2 The last digit should be divisible by 2; i.e. it should be 2, 4, 6, 8 or 04 The number formed by the last 2 digits should be divisible by 48 The number formed by the last 3 digits should be divisible by 816 The number formed by the last 4 digits should be divisible by 163 The sum of its digits should be divisible by 39 The sum of its digits should be divisible by 95 The last digit should be either 5 or 010 The last digit should be 06 It should be divisible by both 2 and 312 It should be divisible by both 3 and 414 It should be divisible by both 2 and 715 It should be divisible by both 3 and 518 It should be divisible by both 2 and 920 It should be divisible by both 4 and 57 (x – 2y) should be divisible by 7*13 (x + 4y) should be divisible by 13*17 (x – 5y) should be divisible by 17*19 (x + 2y) should be divisible by 19** Here, y is the last digit and x is the number formed by the remaining digits. These 4 results canbe derived using the divisibility test for prime numbers.Example: 665 is divisible by 19?

Here y = 5 & x = 66. Then x + 2y = 76 and 76 is divisible by 19. Hence 665 is divisible by 1911 The difference of the sum of the digits in odd positions and thesum of the digits in even positions should be divisible by 11.




Divisibility by Numbers of the form [10n-1] or [10n+1]

Divisibility by 9Let’s Say X=abcdefgh is divisibleby 9 or not?Now, 9 = 101 – 1So, Sum of digits of X is done 1at a time, that is,a + b + c + d + e + f + g + h = PIf P is divisible by 9, then X isalso divisible by 9

Divisibility by 99Let’s Say X=abcdefgh is divisibleby 99 or not?Now, 99 = 102 – 1So, Sum of digits of X is done 2 ata time(starting from last), that is,ab + cd + ef + gh = PIf P is divisible by 99, then X isalso divisible by 99

Divisibility by 999Let’s Say X=abcdefgh isdivisible by 999 or not?Now, 999 = 103 – 1So, Sum of digits of X is done 3at a time(starting from last),that is,ab + cde + fgh = PIf P is divisible by 999, then Xis also divisible by 999Divisibility by 11Let’s Say X=abcdefgh is divisibleby 11 or not?Now, 11 = 101 + 1So, Alternating Subtraction &Summation of digits of X isdone 1 at a time, that is,a - b + c - d + e - f + g - h = PIf P is divisible by 11, then X isalso divisible by 11

Divisibility by 101Let’s Say X=abcdefgh is divisibleby 101 or not?Now, 101 = 102 + 1So, Alternating Subtraction &Summation of digits of X is done2 at a time, that is,ab - cd + ef - gh = PIf P is divisible by 101, then X isalso divisible by 101

Divisibility by 1001Let’s Say X=abcdefgh isdivisible by 1001 or not?Now, 1001 = 103 + 1So, Alternating Subtraction &Summation of digits of X isdone 3 at a time, that is,ab - cde + fgh= PIf P is divisible by 1001, then Xis also divisible by 1001 X is divisible by all factors of 9, 99 & 999. Hence the same divisibility test also work for 3, 11, 27, 37, 41,271 and others X is divisible by all factors of 11, 101 & 1001. Hence the same divisibility test also work for 7,11, 13, 73, 137and others tooDividing Numbers by 5, 25, 12532 / 5----Multiply 32 by 2 = 64-----Put 1 decimal point from the end 64, i.e., 6.4-----Hence 32 /5 = 6.4. Similarly, 133/5 -----133 * 2 =266-----1 decimal point=26.6, hence the answer44 /25-------Multiply 44 by 4 = 176----put decimal point after 2 places from the end, i.e., 1.7690 /125------Multiply 90 by 8 = 720--put decimal point after 3 places from the end, i.e., 0.720




Factors / Divisors of a Given Composite NumberSuppose That a Composite Number ‘n’ is given,Factorize the number n into its prime factors, i.e. n = ax by cz … where a, b, c are prime numbers

Number of Divisors

(x + 1)(y + 1)(z + 1)……For e.g. : 12 = 22 x 31Number of divisors= (2 + 1)x(1 + 1) = 3 x 2 = 6Number of Even & Odd Divisors

Number of even divisors = (x)(y + 1)(z + 1)……Number of odd divisors = (y + 1)(z + 1)……For e.g. : 12 = 22 x 31Number of even divisors= (2 )x(1 + 1) = 2 x 2 = 4Sum of Divisors

Product of Divisors

Number of ways of expressing a givennumber as a product of two factors

½ (x + 1)(y + 1)(z + 1)… (If n is not perfectsquare)½ [(x + 1)(y + 1)(z + 1)… + 1] (if n is perfectsquare)

Number of ways of expressing a givennumber as a product of two Co-Primes

For Eg : 120= 23 x 3 x 5It has 3 distinct prime factors, 2, 3 & 5 (i.e. m =3).So, No of ways 120 can be written as productof two co-primes = [2] = 4Number of sets of factors which are co-

prime to each other

(x+1)(y+1)(z+1)…..-1+ (xy + yz + zx + …..)For Eg : 12 = 22 x 31Number Sets of Co-prime factors= (2+1)(1+1) - 1 + (2 x 1)= 5 + 2 = 7




REMAINDER THEOREMGround RulesRemainder [(a x b) / c]= Remainder [a / c] x Remainder [b / c]Remainder [(a + b) / c]= Remainder [a / c] + Remainder [b / c]Remainder [(a - b) / c]= Remainder [a / c] - Remainder [b / c]

Example:What is the remainder when 1998 x 1999 x2000 is divided by 7?Solution:1998/7 gives Remainder 31999/7 gives Remainder 42000/7 gives Remainder 5So, Final Remainder = 3x4x5 / 7 = 60 /7 = 4Example:What is the remainder when 3232 is divided by9?Solution3232 = (25) * 32 = (2160)Now 26 gives remainder 1 when divided by 9(2160) = 2156 x 242156 gives remainder 1 and 24 gives remainder 7when divided by 9 respectivelyHence final remainder = 7

Case of Negative RemaindersSuppose you have to find remainder when 752 isdivided by 2402?

74 = 2401 will give you remainder (-1) whendivided by 2402. So, we can say that the twice ofthis power will give you remainder (+1), i.e. 78when divided by 2402 will give you remainder(+1).752 = 748 * 74………....748 will give remainder (+1) &74 will be left with remainder 2401 when dividedby 2402. Hence, Final Remainder = 2401Dividend is of the form an + bn / an - bn

Rule 1: an + bn is divisible by a + b when n is ODDRule 2: an - bn is divisible by a + b when n is EVENRule 3: an - bn is ALWAYS divisible by a - bExample:55552222 + 22225555 divided by 7 gives remainder?Solution: 5555 & 2222 gives remainder 4 and 3respectively when divided by7.55552222 + 22225555 divided by7= 42222 + 35555 divided by7= (42) * 1111 + (35) * 1111 divided by7Now it is of the form an + bn when n is ODDSo, the expression will be divisible by 42 + 35 = 16+ 243 = 259, which when divided by 7 givesremainder 0. Hence, Final remainder = 0Fermat’s Remainder TheoremLet p be a prime number and a be a number that isco-prime to p. Then, the remainder obtained whenap-1 is divided by p is 1.Example: Find the remainder when 2100 + 2 isdivided by 101?Solution: 101 is a prime number and 2 is co-primeto 101. 2100 = 2101-1 = This when divided by 101gives remainder 1 according to Fermat’s Theorem.Also, 2 when divided by 101 give 2 as remainder.Hence, Final Remainder = 1 + 2 = 3




Wilson’s Remainder TheoremIf p is a prime number, then (p-1)! + 1 is divisibleby p.Example: Find the remainder when 16! Is dividedby 17?Solution: 17 is a prime number.16! = (16! + 1) – 1 = [(17-1!) +1] – 17+16According to Wilson’s Theorem [(17-1!) +1] isdivisible by 17.Hence, final remainder = 16

Finding the UNIT’S DIGIT

To calculate units digit of where x is a singledigit number. (To calculate unit’s digit ofnumbers like , etc)

Case 1: When y is not a multiple of 4We find the remainder when y is divided by 4. Lety = 4q + r where r is the remainder when y isdivided by 4, and 0 < r < 4. The unit’s digit of xy isthe unit’s digit of xr.Example: Find the unit’s digit of 4347.Solution: The remainder when 47 is divided by 4is 3. Hence the unit’s digit of 4347 is the unit’s digitof 33=7

Case 2: When y is a multiple of 4Even numbers 2, 4, 6, 8 when raised to powerswhich are multiple of 4 give the units digit as 6.Odd numbers 3, 7, and 9 when raised to powerswhich are multiple of 4 give the units digit as 1.Example: Find the unit’s digit of 2828 – 2424.Solution: We have to find the unit’s digit of 828-424. Since 28 and 24 are both multiples of 4, theunit’s digits of both 828-424 will be 6. Hence theunit’s digit of the difference will be 0.Example: Find the unit’s digit of ((711)13)17.Solution: Again, we find the remainder when thepower is divided by 4. Therefore, we find theremainder when (1113)17 is divided by 4.Now 11 = 12 − 1 ⇒ Remainder [11odd] =Remainder [(−1)odd] = −1 = 3.Therefore, unit’s digit of ((711)13)17 = unit’s digitof 73= 3.Finding Last Two Digits Shortcut TechniquesLet’s classify this topic into Finding- Last two digits of numbers which end in 1 Last two digits of numbers which end in 3, 7and 9 Last two digits of numbers which end in 2 Last two digits of numbers which end in 4, 6and 8Last two digits of numbers which end in 1Let's start with few examples.Example: What are the last two digits of 3167?Solution: The unit’s digit is equal to one. Multiplythe tens digit of the number (3 here) with the lastdigit of the exponent (7 here) to get the tens digit.Last two digits are 11.

will always give a remainder of 1for all natural numbers a and n

( ) will always give a remainder of 1when n is EVEN and reminder of a whenn is ODD

The remainder when f(x) = a + bx + cx2 +dx3 +… is divided by [x – a] is f(a) If a number x is divided by n whichleaves a remainder y, then this can berepresented in terms of mod operator as:

x mod n = y




Example: What are the last two digits of 14122?Solution: The last two digits depend only on thelast two digits (41 here). The unit’s digit is equalto one. Multiply the tens digit of the number (4here) with the last digit of the exponent (2 here)to get the tens digit. Last two digits are 81.Last two digits of numbers which end in 3, 7and 9We need to first convert the number such that itgives one as the last digit and then use the sametechnique as mentioned above.Power Cycle of 3 - 3, 9, 7, 1Power Cycle of 7 - 7, 9, 3, 1Power Cycle of 9 - 9, 1So we see that 1 comes as the last digit in thepower cycles of all these three numbers.Therefore if we convert these numbers into theformat of 34k, 74k, 92k format then the last digitbecomes 1 and we can apply our method.Let’s understand this with an example:Example: Find the last two digits of 33288.Solution: 33288 = (334)72.Now let’s find the last two digits of 334 .334 = (332 x 332)Last two digits of (332 x 332) = (89 x 89)Last two digits of 172 that is 89 will be same for332,672, 832,1172 as well ( 332,672 are 50±17 and832,1172 are 100±17 ) .Last two digits of (89 x 89) = 21Same funda as above. Last two digits of 112, 392,612, 892, 1112 will be the same.So now, 33288 = (334 )72 = (21)72

Last two digits of (21)72 = 41 by the method usedin the previous example.(The unit’s digit is equal to one. Multiply the tensdigit of the number (2 here) with the last digit ofthe exponent (2 here) to get the tens digit. Lasttwo digits are 41.)Last two digits of numbers which end in 2Last two digits of (210)even are always 76.Last two digits of (210)odd are always 24.We just need to remember the above two factsand values of 21, 22 up to 210. That’s it!We can apply this for any number in the form 2n.Here’s an example-Example: Find the last two digits of 2543.Solution:2543 = (210)54 x 23Last two digits of (210 )54 = 76Last two digits of 23 = 08Last two digits of 2543 = 76 x 8 = 08Last two digits of numbers which end in 4, 6and 8Now those numbers which are not in the form of2n can be broken down into the form 2n ´ oddnumber. We can find the last two digits of boththe parts separately. Here are some examples:Example: Find the last two digits of 56283Solution: 56283 = ( 23 x 7 )283 = 2849 x 7283= (210)84 x 29 x (74)70 x 73Last two digits = 76 x 12 x 01 x 43 = 16




Number of Exponents / Highest Power /Number of ZeroesLet’s discuss here some basic funda related tofactorials, which form the basis of a largenumber of problems of many topics. Rememberingthe following factorials can be useful-0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6!= 720; 7! = 5040The highest power of a prime number p in n! isgiven by the sum of quotients obtained bysuccessive division of n by p. Let’s understand thisby few examples:Example 1: What is the highest power of 7 that canexactly divide 780! ?Solution: [780/7] = 111; [111/7] = 15;[15/7] = 2;The highest power of 7 = 111 + 15 + 2 = 128Example 2: Find the exponent of 25 in 250!Solution: Here 25 is not prime thus we cannot usethe above formula directly.However 25 = 52; so we will find the exponent(highest power) of 5 in 250! and then divide theexponent value by the power of 5 which is 2 hereto get the exponent of 25.[250/5] = 50; [50/5] = 10; [10/5] = 2;The exponent of 5 is 62.Thus exponent of 25 is 62/2 = 31.Example 3: What is the highest power of 6 thatdivides 134!Solution: As 6 is not a prime number, we willdivide it into its prime factors. 3 is the biggerprime, so its power will be the limiting factor.Hence, we need to find out the power of 3 in 134!

[134/3] = 44; [44/3] = 14; [14/3] = 4; [4/3] = 1;Thus the highest power of 3 in 134! = 44 + 14 + 4 +1 = 63Example 4: How many zeroes are present in (atthe end of) 43!Solution: Number of ending zeroes is the numberof times n! is divisible by 10 or in other words, thehighest power of 10 that divides n!. 10 is not aprime number and its prime factors are 2 and 5. 5being the bigger prime, it’s power becomes thelimiting factor.[43/5] = 8; [9/5] = 1;Thus, 43! contains 9 multiples of 5; hence it willhave 9 zeroes trailing it.Facts about Factorial

Product of n consecutive natural numbers isdivisible by n!11 * 12 * 13 * 14 is completely divisible by4!Consider 1 * 2 * 3 * 4 * 5 * 6 . as per above result1 * 2 * 3 * 4 * 5 * 6 is divisible by 6!.Now take (1 * 2 * 3 ) * ( 4 * 5 * 6 ) , each of this isdivisible by 3! The product of all odd (even) integers up to anodd (even) positive integer n is called the

double factorial of nFor Example: 9!! = 1 * 3 * 5 * 7 * 9 =945

8!! = 2 * 4 * 6 * 8 = 384

A factorion is a natural number that equals thesum of the factorials of its decimaldigits.For example, 145 is a factorion because 1! + 4! +5! = 1 + 24 + 120 =145.




Note: There are just four factorions (in base 10)and they are 1, 2, 145 and 40585Digital Sum of a NumberIf we add up the digits of a number until there isonly one number left, we have found what is calledthe digital sum.Example: What is the digital sum of 78562?Solution: Digital Sum of 78562 = 7+8+5+6+2 = 28= 2+8 = 10 = 1+0 = 1Also the digital sum of the product of two numbersis equal to the product of the digital sums of thetwo numbers. Let’s check this with an example-The product of 795 and 36 is 28620.Digital sum of 795 = 7+9+5 = 21 = 2+1 = 3Digital sum of 36 = 3+6 = 9Product of these digital sums= 9x3= 27= 2+7= 9Digital sum of 28620= 2+8+6+2+0= 18= 1+8= 9Applications of Digital SumThis concept can be used almost everywhere! Thiscan be helpful in the rapid checking of calculationswhile multiplying numbers. It greatly simplifiesthe calculation of Simple and Compound interestwhich we will discuss in a later unit. We can usethis to check whether a number is a perfect squareor not as the digital sum of perfect squares willalways be 1, 4, 7 or 9.

1+2+3+ . . . . + 10 = 5511 + 12 + . . . . + 20 = 15521 + 22 + . . . . + 30 = 25531 + 32 + . . . . + 40 = 355Example: Find the 281st term of the sequence.ABBCCCDDDDEEEEE…Solution: Here, A is appearing for 1 time, B for 2times, C for 3 times and so on…….Now from above mentioned series we get that1+2+3+4+......+20 = 55 + 155 = 210;210 + 21 + 22 + 23 = 276; and we need to find 281st,so 276 + 24 = 300 which covers the 281st part.Hence, 24th letter will be 281st term of sequence.Hence, the answer is “X”.

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01 Number System Aptitude Trick-Book AptiDude

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