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Factor of the product of the first 24 positive integers

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by Matt@VeritasPrep » Wed Feb 19, 2014 5:06 pm
To see what's happening, let's start with smaller sets.

Suppose I have the product of the first five positive integers:

5 * 4 * 3 * 2 * 1

This divides by 10 because of the two factors in red.

Now suppose I have the product of the five ten positive integers:

10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

This will divide by 100 because of the three factors in red.

So now we have the idea: every time we can find a 10 (either a proper power of 10, or a pair of a multiple of 2 and a multiple of 5) in the product, we can divide by 10.

Looking at the product of the first 24 positive integers, we have four multiples of 5: 5, 10, 15, and 20. Each of these can be paired with an even integer to generate a multiple of 10, so 24! is divisible by 10�.
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by [email protected] » Wed Feb 19, 2014 11:49 pm
Hi Sri,

This question is essentially about "prime factorization", although you don't need to prime factor all of the numbers to get to the solution.

Prime factorization is the math concept that all positive integers can be "broken down" into a series of prime numbers multiplied together (unless the number is already prime).

So, the number 10 = 2x5, which the number 13 is just 13.

This question tells us to take the product of the first 24 positive integers, so we're supposed to think about 1x2x3x....24. There's NO WAY that the GMAT would actually ask us to do that math, so we have to focus on the specific question that's asked: 10^N IS a factor of that product, so what's the biggest number that's possible for N?

Since 10 = 2x5, we're looking for 2s and 5s in the product of that big number:

5 = 5
10 = 5x2
15 = 5x3
20 = 5x4

There are LOTS of 2s in the given product (in every even number; sometimes there's more than 1...e.g. 8 = 2x2x2), but there are only four 5s. Since there are only four 5s, then we're going to end up with 10^4, so the greatest possible value of N is 4.

Final Answer: D

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