Problem Solving

If m is a product of all integers from 1 to 40 inclusive, what is the greatest integer P for which 10^P is a factor of M?
a) 7
B) 8
c) 9
D) 10
E) 11

Every explanation I receive and read of how to do this problem goes in one ear and out the other so I am curious to see how everybody here makes it work. Looking forward to your replies.

The answer is C

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.

I am confused, because I've tested for p [0< p <= 33]
p can be 33 hence 10 and 11
why p=9?

P isn't the number that you're dividing by. P is the power of ten that's the largest you can divide 10! by.

Laura GMAT Tutor wrote:P isn't the number that you're dividing by. P is the power of ten that's the largest you can divide 10! by.

40!=815915283247898000000000000000000000000000000000,00


p(max)=33

815915283247898000000000000000000000000000000000,00 : 10^33 = 815915283247898

First, is it feasible on test day to actually work out 40! ? Clearly that approach isn't correct in terms of methodology. If you get in the habit of solving things in ways that are completely not feasible in under 2.5 minutes maximum, then you're harming yourself in the long run.

Second, I'm not sure where you're getting your numbers. 40! is 815915283247897734345611269596115894272000000000. As you can see, that only has 9 zeros in it.

I still think that you're misunderstanding the premise of the problem, but I don't know what in particular you think the question is asking for.

Laura GMAT Tutor wrote:First, is it feasible on test day to actually work out 40! ? Clearly that approach isn't correct in terms of methodology. If you get in the habit of solving things in ways that are completely not feasible in under 2.5 minutes maximum, then you're harming yourself in the long run.

Second, I'm not sure where you're getting your numbers. 40! is 815915283247897734345611269596115894272000000000. As you can see, that only has 9 zeros in it.

I still think that you're misunderstanding the premise of the problem, but I don't know what in particular you think the question is asking for.

ok i was hand calculating using some math wiz
now excel calc 815915283247898 plus 33 zeros

i don't know may be excel mistake

But it's all irrelevant. You can't use a calculator to work out 40! on test day. You can't use excel or an online calculator or anything. And you don't have time to do it by hand. The value of 40! is irrelevant. What matters is your solving methodology, and it is COMPLETELY NOT FEASIBLE to do this any other way than by using factors.

If you want to get better at the test, you have to learn new methods. And factoring is a vitally important method.

Laura GMAT Tutor wrote:But it's all irrelevant. You can't use a calculator to work out 40! on test day. You can't use excel or an online calculator or anything. And you don't have time to do it by hand. The value of 40! is irrelevant. What matters is your solving methodology, and it is COMPLETELY NOT FEASIBLE to do this any other way than by using factors.

If you want to get better at the test, you have to learn new methods. And factoring is a vitally important method.

why i get the result of 10^33 when I use excel, that's what interests me

Laura, as for the solution of this problem, actually I was verse playing with prime factorization firstly - but it did not work.

my worry is about why we get the power of 33 for 10 by using excel?

calculator won't give you the factor of 40 anyway... it's just a way too long for its memory, might be overflown

Thanks Laura; she supplied me with online calculator at https://www.calculatorsoup.com/calculato ... orials.php

excel spreadsheet values are not valid for the big amount entries...

I have figured out what you're doing wrong, Night Reader, and just in case anyone else is wondering I'll explain.

Our job is to figure out how many 10s we can make from the list of factors, but it has to be AT THE SAME TIME. We aren't making new combinations of 2s and 5s to make new 10s. It's how many tens can we make using only the number of 5s and 2s that we have.

I can't do this for 40! so I'll just do it for 20!.

(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19)(20) =
=(2)(3)(2^2)(5)(2x3)(7)(2^3)(3^2)(2x5)(11)(2^2x3)(13)(2x7)(3x5)(2^4)(17)(2x3^2)(19)(2x2x5) =
=(2^18)(3^8)(5^4)(7^2)(11)(13)(17)(19)
=(2^18)(5^4)(3^8)(7^2)(11)(13)(17)(19)
=(2^14)(5^4)(2^4)(3^8)(7^2)(11)(13)(17)(19)
=(10^4)(2^14)(3^8)(7^2)(11)(13)(17)(19)

so how many times can you divide that number by 10? 4 times.

If you did the same thing all the way to 40 you'd get 9 tens.

i normally don't add to the threads here, but there is a much easier way to solve this problem. It involves trailing zeros. to find how many factors of 10 there are, simple take N, which in this case is 40 and divide it by 5. then square 5, 25, and divide 40 by 25. 25 goes into 40 one time. so, 5 goes into 40 8 times and 25 goes into 40 1 giving 8+1=9 factors of 10. This is true for all numbers. Check out the "trailing zeros" in the search engine and their is a better explanation.