Difference between revisions of "2013 AMC 12B Problems/Problem 23"
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<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D}}\ 20\qquad\textbf{(E)}\ 25 </math> | <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D}}\ 20\qquad\textbf{(E)}\ 25 </math> | ||
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| + | ==Solution== | ||
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| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}} | ||
Revision as of 17:16, 22 February 2013
Problem
Bernardo chooses a three-digit positive integer 
and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer 
. For example, if 
, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum 
. For how many choices of are the two rightmost digits of , in order, the same as those of 
?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D}}\ 20\qquad\textbf{(E)}\ 25$ (Error compiling LaTeX. Unknown error_msg)
Solution
See also
| 2013 AMC 12B (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 | |
