solns4 - Mathematics 55 Summer 2006 SolutionsProblem Set 4...

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Mathematics 55 — Summer 2006 Solutions–Problem Set 4 3.4.6 Solution : The number of zeros at the end of 100! is just the number of times that 10 divides 100!. However, 10 is not prime, so it will not work just to take all the numbers between 1 and 100 that 10 divides. What we really need here is the Fundamental Theorem of Arithmetic , which says that every number has a unique factorization into prime numbers . Since 10 = 2 · 5, we just need to count the factors of 5 and those of 2 in 100!. Since 5’s are easier to count, start there. Then we’ll see that we don’t really need to count all of the 2’s. Exactly 20 of the numbers between 1 and 100 are divisible by 5 (every 5th number). However, some of them are divisible by two 5’s, i.e. by 25. There are 4 of those, namely 25,50,75 and 100. We can count one 5 for each of the 20 numbers that 5 divides, but we need to count 4 more 5’s for the numbers that 25 divides. Since 5 3 = 125, there are no numbers between 1 and 100 which are divisible by more than two 5s. Therefore, the largest power of 5 which divides 100! is 24. There are a lot of 2’s that divide 100!. Since half of the numbers between 1 and 100 are even, there are certainly more than 50. But we only have 24 5’s to pair with these 2’s in order to make 10’s. So as long as we know that we have at least 24 (fortunately 50
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solns4 - Mathematics 55 Summer 2006 SolutionsProblem Set 4...

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