Difference between revisions of "2000 AMC 12 Problems/Problem 23"
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== Problem == | == Problem == | ||
− | 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? | + | 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? |
− | <math>\ | + | <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 == | ||
− | The product of the numbers | + | 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>. | ||
− | + | ||
+ | == Video Solution == | ||
+ | https://youtu.be/9vP9Efu18Jk | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2000|num-b=22|num-a=24}} | {{AMC12 box|year=2000|num-b=22|num-a=24}} | ||
− | [[Category:Introductory | + | [[Category:Introductory Probability Problems]] |
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:26, 24 October 2021
Contents
Problem
Professor Gamble buys a lottery ticket, which requires that he pick six different integers from through
, 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?
Solution
The product of the numbers has to be a power of in order to have an integer base ten logarithm. Thus all of the numbers must be in the form
. Listing out such numbers from
to
, we find
are the only such numbers. Immediately it should be noticed that there are a larger number of powers of
than of
. Since a number in the form of
must have the same number of
s and
s in its factorization, we require larger powers of
than we do of
. To see this, for each number subtract the power of
from the power of
. This yields
, and indeed the only non-positive terms are
. Since there are only two zeros, the largest number that Professor Gamble could have picked would be
.
Thus Gamble picks numbers which fit , with the first four having already been determined to be
. The choices for the
include
and the choices for the
include
. Together these give four possible tickets, which makes Professor Gamble’s probability
.
Video Solution
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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