Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎See also)
Line 15: Line 15:
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 20:36, 3 July 2013

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?

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

Solution

The product of the numbers have 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 $2^m5^m$. 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 $5$ of those 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 $1/4\ \mathrm{(B)}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_23&oldid=53572"