2002 AMC 10P Problems/Problem 22
Contents
[hide]Problem
In how many zeroes does the number end?
Solution 1
We can solve this problem with an application of Legendre's Formula.
We know that there will be an abundance of factors of compared to factors of so finding the amount of factors of is equivalent to finding how many factors of there are, which is equivalent to how many zeroes there are at the end of the number. Additionally, squaring a number will multiply the exponent of each factor by Therefore, we plug in and then plug in and and multiply by in:
As such,
or alternatively,
Similarly,
or alternatively,
In any case, our answer is
Solution 2
In case we have forgotten Legendre's formula or haven't learned it, this solution is equally viable. With similar reasoning to solution 1, all we need to find is the amount of multiples of in the problem.
Cancel from the top and bottom of the fraction. We get We can set a bijection between the two sets of terms with a multiple of Let be a number from that is a multiple of Its corresponding multiple of from will be For clarity, will group with will group with will group with Our goal is to find number(s) where for all will result in a multiple of Simplifying this equation, a multiple of We can conclude that is the only solution to this equation since is not divisible by any other number ending in a or (we can write the factors of to confirm this; and are the only numbers that do not end in or because all multiples of end in or ). A quick check reveals which has multiples of and which has multiples of Thus, is the only instance where there is an extra multiple of at the top, meaning our answer is
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
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