if d is a positive integer, f is the product of the first 30 positive integers,
what is the value of d?

(1) 10^d is a factor of f
(2) d > 6

??
I know how to answer this by counting 5,2 and 10's. I'm wondering if theres a quicker, smarter way..

Thanks!

Is the only word missing from your transcription the "and" (I assume) that should be after the first comma? I'll assume so - please double check. If something else is missing or entered incorrectly, please correct it!

To answer your question: you sort of have to do some counting of 5's and 2's (well, 5's anyway). But you can do some thoughtful analysis to really shortcut the work.

d = + int
f = 30! = 30*29*28*...*3*2*1
value of d = ?

Start with statement 2. This doesn't tell us one value of d, so elim. B and D.

Statement 1: 10^d is a factor of f. This isn't going to be sufficient. If you're not sure why try the easiest possible positive integers. Is 10^1 = 10 a factor of f? Yes, so 1 is a possible value for d. Is 10^2 = 100 a possible factor of f? Yes, so 2 is a possible value for d. I just found 2 possible values for d. Elim. A. Only C and E are left.

d is a pos int (given in stem) and is greater than 6 (statement 2). Smallest possibility, then, is 7. If d is anything greater than 7, then 7 will work too (eg, if d actually is 8, then 7 would also satisfy both statements and we wouldn't be able to tell, just from the statements, whether d is 7 or 8). So it's either 7, exactly, which is sufficient, or it's something greater than 7, which is not sufficient.

So how many 10's are in f?
write down the numbers that contain 2s and 5s (only those)
30*28*26*25*24*22*20*18*16*15*14*12*10*8*6*5*4*2

Now ask yourself Is my limiting factor going to be 5 or is it going to be 2?

It's going to be 5 because there are many more 2's up there. So circle the numbers that contain 5's:
30, 25, 20, 15, 10, 5
How many 5's do you have? Seven 5's (don't forget - 25 has two 5's!), so you can make seven 10's. That's it. Answer is C.

Hi instructors,
Quick question on this problem-Could please clarify by what you mean when you state, "Now ask yourself Is my limiting factor going to be 5 or is it going to be 2? It's going to be 5 because there are many more 2's up there." Why have you chosen 5 over 2 is there a difference? If a similar problem arises do I just eliminate the number that has more numbers?

Your input is greatly appreciated...

By "limiting factor" Stacey means "which is least common or likely." Think of it this way: there are many more multiples of 2 than there are multiples of 5. In probability terms, a number is more likely to be even than to be a multiple of 5. In divisibility terms, take some large number that is divisible by both 2 and 5, and it is likely to have more factors of 2 than 5.

For example: 400 = 4*10*10 = (2*2)(2*5)(2*5) = (2^4)(5^2).

I know, numbers with more factors of 5 than factors of 2 exist...this is just a bet we make to ease the computation.

In general, the larger the factor, the less likely it is to divide evenly into a number. The larger the factor, the more of a "limiting factor" it is.