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The Cell Phone Dropping Problem Japheth Wood Philip B. Yasskin Bard College Texas A&M University August 7, 2015

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The Cell Phone Dropping Problem

Japheth Wood Philip B. Yasskin

Bard College Texas A&M University

August 7, 2015

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The Cell Phone Dropping Problem

You work for a cell phone com-pany which has just invented anew cell phone protector andwants to advertise that it canbe dropped from the N-th floorwithout breaking.

If you are given 1 or 2 phones and a 100 story building, how do youguarantee you know the highest floor it won’t break with thesmallest number of trial drops?

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With 1 phone: Drop from every floor, may require 100 drops.

With 2 phones, you can try:

I Drop from every 2nd floor. 50 + 1 = 51 drops

I Drop from every 50th floor. 2 + 49 = 51 drops

I Drop from every 25th floor. 4 + 24 = 28 drops

I Drop from every kth floor. drops(k) = b100/kc+ k − 1

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Try it: drops(k) = b100/kc+ k − 1

k drops(k)

7 b100/7c+ 7− 1 = 208 b100/8c+ 8− 1 = 199 b100/9c+ 9− 1 = 19

10 b100/10c+ 10− 1 = 1911 b100/11c+ 11− 1 = 1912 b100/12c+ 12− 1 = 1913 b100/13c+ 13− 1 = 1914 b100/14c+ 14− 1 = 20

Or, for those who know some calculus, if f (k) = 100/k + k − 1,then f ′(k) = −100/k2 + 1. Critical point f ′(k) = 0 when k = 10.

So the minimum seems to be f (10) = 19, but this is wrong.What was the incorrect assumption?

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SEE-Math

Dr. Yasskin was given thisproblem by a middle schoolstudent at SEE-Math whohad learned it at MathPath.Yasskin has used it success-fully at the TAMU Math Cir-cle with high school studentsand at a session of 30 David-son Young Scholars.

Summer Educational Enrichment in Math MathPathsee-math.math.tamu.edu www.mathpath.org

TAMU Math Circle Davidson Young Scholars

mathcircle.tamu.edu www.davidsongifted.org

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SPMPS/BEAM

Used as a Challenge Problem at SPMPS. Solving it earns amovie night at the end of the week for the whole program.

Summer Program in Mathematical Problem Solving (SPMPS)

www.artofproblemsolving.org/spmps

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12. Marbles

10/3/13 12. Marbles, Science, Technology and Engineering Students, La Trobe University

www.latrobe.edu.au/students/scitecheng/department-resources/mathematics-and-statistics/puzzle-of-the-week/12.-marbles 1/1

Science, Technology and Engineering Students

12. MarblesYou are standing in front of a building with 100 levels, and

you are given two glass marbles.

If you drop a glass marble out of the window on level 100, it

will break.

Your job is to determine the lowest level at which a glassmarble will break if dropped out of the window.

The simple­minded method is to drop a marble first from

level 1, then from level 2, and so on, testing each level until

the marble breaks. This method will require a maximum of 99

tests.

Since you have two marbles, you can use a more efficient method ­ if one marble breaks, you can

continue working with the other.

Find the most efficient method for using the two marbles to determine the lowest breaking­level.

In other words, find the method that has the smallest possible maximum number of tests. What is

this number?

© Copyright 2013 La Trobe University. All rights reserved.CRICOS Provider Code: VIC 00115M, NSW 02218K

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10/3/13 12. Marbles, Science, Technology and Engineering Students, La Trobe University

www.latrobe.edu.au/students/scitecheng/department-resources/mathematics-and-statistics/puzzle-of-the-week/12.-marbles 1/1

Science, Technology and Engineering Students

12. MarblesYou are standing in front of a building with 100 levels, and

you are given two glass marbles.

If you drop a glass marble out of the window on level 100, it

will break.

Your job is to determine the lowest level at which a glassmarble will break if dropped out of the window.

The simple­minded method is to drop a marble first from

level 1, then from level 2, and so on, testing each level until

the marble breaks. This method will require a maximum of 99

tests.

Since you have two marbles, you can use a more efficient method ­ if one marble breaks, you can

continue working with the other.

Find the most efficient method for using the two marbles to determine the lowest breaking­level.

In other words, find the method that has the smallest possible maximum number of tests. What is

this number?

© Copyright 2013 La Trobe University. All rights reserved.CRICOS Provider Code: VIC 00115M, NSW 02218K

Print version Close

You are standing in front of a building with 100 lev-els, and you are given two glass marbles. If you dropa glass marble out of the window on level 100, it willbreak.

Your job is to determine the lowest level at which a glass marble willbreak if dropped out of the window. The simple-minded method is todrop a marble first from level 1, then from level 2, and so on, testingeach level until the marble breaks. This method will require a maximumof 99 tests.

Since you have two marbles, you can use a more efficient method if onemarble breaks, you can continue working with the other.

Find the most efficient method for using the two marbles to determinethe lowest breaking level. In other words, find the method that has thesmallest possible maximum number of tests.

What is this number?

La Trobe University, Mathematics and Statistics Puzzle of the Week (2013): www.latrobe.edu.au

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Testing Ostrich Eggs

Mathematical Mind-Benders, Peter Winkler

In preparation for an ad campaign, the FlightlessOstrich Farm needs to test its eggs for durabil-ity. The world standard for egg-hardness callsfor rating an egg according to the highest floorof the Empire State Building from which the eggcan be dropped without breaking.

Flightless’s official tester, Oskar, realizes that if he takes only oneegg along on his trip to New York, he’ll need to drop it from(potentially) every one of the building’s 101 floors, starting withthe first, to determine its rating.

How many drops does he need in the worst cast, if he takes twoeggs?

Mathematical Mind-Benders by Peter Winkler, 2007. A K Peters/CRC Press

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INFORMS Transactions on Education

The Joy of Egg-Dropping in Braunschweig and Hong KongMoshe Sniedovich, University Of Melbourne

This article was downloaded by: [67.243.51.210] On: 30 July 2015, At: 19:54Publisher: Institute for Operations Research and the Management Sciences (INFORMS)INFORMS is located in Maryland, USA

INFORMS Transactions on Education

Publication details, including instructions for authors and subscription information:http://pubsonline.informs.org

OR/MS Games: 4. The Joy of Egg-Dropping inBraunschweig and Hong KongMoshe Sniedovich,

To cite this article:Moshe Sniedovich, (2003) OR/MS Games: 4. The Joy of Egg-Dropping in Braunschweig and Hong Kong. INFORMS Transactionson Education 4(1):48-64. http://dx.doi.org/10.1287/ited.4.1.48

Full terms and conditions of use: http://pubsonline.informs.org/page/terms-and-conditions

This article may be used only for the purposes of research, teaching, and/or private study. Commercial useor systematic downloading (by robots or other automatic processes) is prohibited without explicit Publisherapproval, unless otherwise noted. For more information, contact [email protected].

The Publisher does not warrant or guarantee the article’s accuracy, completeness, merchantability, fitnessfor a particular purpose, or non-infringement. Descriptions of, or references to, products or publications, orinclusion of an advertisement in this article, neither constitutes nor implies a guarantee, endorsement, orsupport of claims made of that product, publication, or service.

Copyright © 2003, INFORMS

Please scroll down for article—it is on subsequent pages

INFORMS is the largest professional society in the world for professionals in the fields of operations research, managementscience, and analytics.For more information on INFORMS, its publications, membership, or meetings visit http://www.informs.org

I Operations Research

I Management Sciences

I Dynamic Programming

I Worst Case scenario

I Expected Value scenario

Moshe Sniedovich, (2003) OR/MS Games: 4. The Joy of Egg-Dropping in Braunschweig and Hong Kong.INFORMS Transactions on Education 4(1):48–64.

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166. An Egg-Drop Experiment (p.53)

Which Way Did the Bicycle Go?

Suppose that we wish to know which windows in a 36-storybuilding are safe to drop eggs from, and which will cause the eggsto break on landing. We make a few assumptions:

I An egg that survives a fall can be used again.

I A broken egg must be discarded.

I The effect of a fall is the same for all eggs.

I If an egg breaks when dropped, then it would break if dropped from a higherwindow.

I If an egg survives a fall then it would survive a shorter fall.

I It is not ruled out that the first-floor windows break eggs, nor is it ruled outthat the 36th-floor windows do not cause an egg to break.

If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried outin only one way. Drop the egg from the first-floor window; if it survives, drop it from the second floor window.Continue upward until it breaks. In the worst case, this method may require 36 droppings. Suppose 2 eggs areavailable. What is the least number of egg-droppings that is guaranteed to work in all cases?

Which Way Did the Bicycle Go? by Joseph Konhauser, Dan Velleman, Stan Wagon. (1996). MAA DolcianiMathematical Expositions

Macalester College Problem of the Week, 1968–1995. (See mathforum.org/wagon for 1995–present).

Stan Wagon’s favorite problems/puzzles: stanwagon.com/wagon/Misc/bestpuzzles.html

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The Solution (part 1): 14 drops is sufficient

105th floor1

104th floor2

102nd floor3

99th floor...

50th floor11

39th floor12

27th floor13

14th floor14

14 + 13 + · · ·+ 2 + 1 = 105 > 100

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The Solution (part 2): 13 drops is not sufficient

1 = phone breaks 0 = phone doesn’t break

1 2 3 4 5 6 7 8 9 10 11 12 13

0 0 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0 0 10 0 0 0 0 1 0 0 0 0 0 10 0 0 0 0 1 0 0 0 0 1

I Each trial yields a string in {1, 0} of length ≤ 13.

I No symbol can appear to the right of a second 1.

I No string can be the prefix of another string.

I There must be at least 100 strings.

(13

2

)+

(13

1

)+

(13

0

)= 78 + 13 + 1 = 92 < 100

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With 14 Drops

(14

2

)+

(14

1

)+

(14

0

)= 91 + 14 + 1 = 106 > 100

Why does(N

2

)+(N

1

)+(N

0

)always add up to one more than a

triangular number?

What is the solution to the Expected Value scenario?