2006 AIME I Problems/Problem 4
Contents
Problem
Let be the number of consecutive 's at the right end of the decimal representation of the product Find the remainder when is divided by .
Solution
A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: of the numbers have a factor of . have a factor of . have a factor of . And so on. This gives us an initial count of . Summing this arithmetic series of terms, we get . However, we have neglected some powers of - every term for has an additional power of dividing it, for extra; every n! for has one more in addition to that, for a total of extra; and similarly there are extra from those larger than and extra from . Thus, our final total is , and the answer is .
Solution 2 (Legendre's Formula )
This problem can be easily solved using Legendre's Formula:
First is to account for all of the factorials that are greater than or , then the factorials that are greater than or . Since we do not need to account for it.
This gives and respectively. Adding all of these terms up gives 1124 or .
~PeterDoesPhysics
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=413
~ pi_is_3.14
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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