2010 AMC 12A Problems/Problem 23
Problem
The number obtained from the last two nonzero digits of is equal to . What is ?
Solution
We will use the fact that for any integer ,
First, we find that the number of factors of in is equal to . Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .
If we divide by by taking out all the factors of in , we can write as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .
The number can be grouped as follows:
Using the identity at the beginning of the solution, we can reduce to
Using the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore .
Finally, combining with the fact that yields .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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All AMC 12 Problems and Solutions |