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Difference between revisions of "2000 AMC 12 Problems/Problem 23"

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== Problem ==
 
== Problem ==
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Professor Gamble buys a lottery ticket, which requires that he pick six different integers from <math>1</math> through <math>46</math>, inclusive. He chooses his numbers so that the sum of the base-ten [[logarithm]]s of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the [[probability]] that Professor Gamble holds the winning ticket?
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Professor Gamble buys a lottery ticket, which requires that he pick six different integers from <math>1</math> through <math>46</math>, inclusive. He chooses his numbers so that the sum of the base-ten [[logarithm]]s of his six numbers is an [[integer]]. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the [[probability]] that Professor Gamble holds the winning ticket?
  
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<math>\text {(A)}\ 1/5 \qquad \text {(B)}\ 1/4 \qquad \text {(C)}\ 1/3 \qquad \text {(D)}\ 1/2 \qquad \text {(E)}\ 1</math>
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<math>\textbf {(A)}\ 1/5 \qquad \textbf {(B)}\ 1/4 \qquad \textbf {(C)}\ 1/3 \qquad \textbf {(D)}\ 1/2 \qquad \textbf {(E)}\ 1</math>
  
 
== Solution ==
 
== Solution ==
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The product of the numbers have to be a power of <math>10</math> in order to have an integer base ten logarithm. Thus all of the numbers must be in the form <math>2^m5^m</math>. Listing out such numbers from <math>1</math> to <math>46</math>, we find <math>1,2,4,5,8,10,16,20,25,32,40</math> are the only such numbers. Immediately it should be noticed that there are a larger number of powers of <math>2</math> than of <math>5</math>. Since a number in the form of <math>10^k</math> must have the same number of <math>2</math>s and <math>5</math>s in its factorization, we require larger powers of <math>5</math> of those of <math>2</math>. To see this, for each number subtract the power of <math>5</math> from the power of <math>2</math>. This yields <math>0,1,2,-1,3,0,4,1,-2,5,2</math>, and indeed the only non-positive terms are <math>0,0,-1,-2</math>. Since there are only two zeros, the largest number that Professor Gamble could have picked would be <math>2</math>.  
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The product of the numbers has to be a power of <math>10</math> in order to have an integer base ten logarithm. Thus all of the numbers must be in the form <math>2^m5^n</math>. Listing out such numbers from <math>1</math> to <math>46</math>, we find <math>1,2,4,5,8,10,16,20,25,32,40</math> are the only such numbers. Immediately it should be noticed that there are a larger number of powers of <math>2</math> than of <math>5</math>. Since a number in the form of <math>10^k</math> must have the same number of <math>2</math>s and <math>5</math>s in its factorization, we require larger powers of <math>5</math> than we do of <math>2</math>. To see this, for each number subtract the power of <math>5</math> from the power of <math>2</math>. This yields <math>0,1,2,-1,3,0,4,1,-2,5,2</math>, and indeed the only non-positive terms are <math>0,0,-1,-2</math>. Since there are only two zeros, the largest number that Professor Gamble could have picked would be <math>2</math>.  
  
 
Thus Gamble picks numbers which fit <math>-2 + -1 + 0 + 0 + 1 + 2</math>, with the first four having already been determined to be <math>\{25,5,1,10\}</math>. The choices for the <math>1</math> include <math>\{2,20\}</math> and the choices for the <math>2</math> include <math>\{4,40\}</math>. Together these give four possible tickets, which makes Professor Gamble’s probability <math>1/4\ \mathrm{(B)}</math>.
 
Thus Gamble picks numbers which fit <math>-2 + -1 + 0 + 0 + 1 + 2</math>, with the first four having already been determined to be <math>\{25,5,1,10\}</math>. The choices for the <math>1</math> include <math>\{2,20\}</math> and the choices for the <math>2</math> include <math>\{4,40\}</math>. Together these give four possible tickets, which makes Professor Gamble’s probability <math>1/4\ \mathrm{(B)}</math>.
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== Video Solution ==
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https://youtu.be/9vP9Efu18Jk
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 21:26, 24 October 2021

Problem

Professor Gamble buys a lottery ticket, which requires that he pick six different integers from $1$ through $46$, inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?

$\textbf {(A)}\ 1/5 \qquad \textbf {(B)}\ 1/4 \qquad \textbf {(C)}\ 1/3 \qquad \textbf {(D)}\ 1/2 \qquad \textbf {(E)}\ 1$

Solution

The product of the numbers has to be a power of $10$ in order to have an integer base ten logarithm. Thus all of the numbers must be in the form $2^m5^n$. Listing out such numbers from $1$ to $46$, we find $1,2,4,5,8,10,16,20,25,32,40$ are the only such numbers. Immediately it should be noticed that there are a larger number of powers of $2$ than of $5$. Since a number in the form of $10^k$ must have the same number of $2$s and $5$s in its factorization, we require larger powers of $5$ than we do of $2$. To see this, for each number subtract the power of $5$ from the power of $2$. This yields $0,1,2,-1,3,0,4,1,-2,5,2$, and indeed the only non-positive terms are $0,0,-1,-2$. Since there are only two zeros, the largest number that Professor Gamble could have picked would be $2$.

Thus Gamble picks numbers which fit $-2 + -1 + 0 + 0 + 1 + 2$, with the first four having already been determined to be $\{25,5,1,10\}$. The choices for the $1$ include $\{2,20\}$ and the choices for the $2$ include $\{4,40\}$. Together these give four possible tickets, which makes Professor Gamble’s probability $1/4\ \mathrm{(B)}$.

Video Solution

https://youtu.be/9vP9Efu18Jk

See also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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