Difference between revisions of "2010 AMC 12A Problems/Problem 23"
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Where the first line is composed of the numbers in <math>90!</math> that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25. | Where the first line is composed of the numbers in <math>90!</math> that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25. | ||
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Using the identity at the beginning of the solution, we can reduce <math>M</math> to | Using the identity at the beginning of the solution, we can reduce <math>M</math> to | ||
Revision as of 14:55, 31 December 2011
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:
Where the first line is composed of the numbers in that aren't multiples of five, the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.
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 |
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 |