2010 AMC 10A Problems/Problem 24
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 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AMC 10 Problems and Solutions |