Let's start with the basics.
An integer x will be a factor of an integer y if you can divide y by x and get an integer. This basically means that all the prime factors on the bottom of the fraction will cancel on the top.
For example, 15 is divisible by 5. If we re-write 15 as (5)(3) we can see that in
(5)(3)/ 5
the 5 on the bottom cancels with the 5 on the top and all that's left is the 3 on top.
15 is not divisible by 7 because in
(5)(3)/7
the 7 is "stuck" on the bottom.
So. Back to this problem. If you divide m by some large power of 10, what's the biggest power of 10 you can divide it by?
Let's just start with the product of the integers from 1 to 10.
(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)
Remember, each 10 is a (5)(2).Clearly we can divide that by 10 once because there's an actual 10 in that list of numbers. We can also divide m by 10 a second time because there's an extra 5 and a 2. But that's it. If we try to divide that by 10 one more time, we wouldn't be able to. We've already used up all the 5s.
So, what holds us back from being able to divide by 10 is the number of 5s and 2s we have. 5 is more restrictive since there are plenty of 2s to go around.
So in the product of (1)(2)(3)(4).....(37)(38)(39)(40) we definitely have four 10s already -- one in 10, one in 20 (2x10), one in 30 (3x10) and one in 40 (4x10). Then there's all the tens we can make out of other 5s and 2s. There's a 5, 15, 25, and 35, which is a total of five 5s (25 has two 5s in it). There are plenty of 2s on that list to use as well, so that's another 5 tens we can make, for a total of 9 tens.
You also could, I suppose, write out the entire list, then break the entire list into prime factors. It would just take a long time.
Last edited by
Laura GMAT Tutor on Fri Dec 03, 2010 4:14 pm, edited 1 time in total.
