Difference between revisions of "2000 AMC 12 Problems/Problem 23"
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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 == | ||
Latest revision as of 21:26, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
