Difference between revisions of "2010 AMC 12A Problems/Problem 23"
(→Solution) |
(→Solution) |
||
Line 12: | Line 12: | ||
&=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath> | &=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath> | ||
− | First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>N\pmod{100}</math>. Since there is clearly an excess of factors of 2, we know that <math>N\equiv 0\pmod 4</math>, so it remains to find <math> | + | First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>N\pmod{100}</math>. Since there is clearly an excess of factors of 2, we know that <math>N\equiv 0\pmod 4</math>, so it remains to find <math>N\pmod{25}</math>. |
If we take out all the factors of <math>5</math> in <math>N</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where | If we take out all the factors of <math>5</math> in <math>N</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where |
Revision as of 16:19, 12 February 2010
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 take out all the factors of in
, we can write
as
where
where every factor 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
.