Tough Multiples Question #3

[theboyleman32]

[ceilidh.erickson]

This question is a variation on OG PS #116. The most difficult thing is translating what the question is asking.

"m is the product of all the integers from 1 to 40 inclusive" --> This means that m = 40!

"What is the greatest integer p for which 10^p is a factor of m?" --> If we're being asked to maximize the exponent of 10, then the question is really asking: how many factors of 10 are in 40 factorial? Or in other words, how many times does 10 go into 40 factorial?

We definitely don't want to calculate what 40! is - we just need to count the factors of 10.

But we can't simply count 10, 20, 30, and 40. We also need to consider that 2 x 5 = 10, 15 x 4 = 60, etc. What we really need to do is count the 5's, because every factor of 5 will have a corresponding factor of 2. So, just add them up:
5, 10, 15, 20, 25, 30, 35, 40

It looks like 8, but remember - 25 has two factors of 5 in it, so we really have 9 factors of 5, and thus 9 factors of 10 in 40!.

The answer is D.

[[email protected]]

Hi Tom,

This question comes down to "prime factorization" and finding all the "5s" and "10s" in the first 40 positive integers.

We're asked to largest value of P so that 10^P divides into the product of the first 40 positive integers.

10 = 5x2...so we're looking for all of the 10s that can be "made" in this big product.

Let's look at all the multiples of 5....

5, 10, 15, 20, 25, 30, 35, 40

So we have some obvious 10s

10
20 = 2x10
30 = 3x10
40 = 4x10

We can "make" some more 10s with the numbers that end in 5....

5x(any even) = multiple of 10
15x(any even) = multiple of 10
35x(any even) = multiple of 10

I've separated the 25 for a reason; it creates an EXTRA 0 when multiplied by a multiple of 4...

25x4 = 100 = 10^2

So we have....4 + 3 + 2 = 9 10s.....P = 9

Final Answer:D

GMAT assassins aren't born, they're made,
Rich