vikkimba17 wrote:If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3^k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

This question is really asking us to determine the number of 3's "hiding" in the prime factorization of p.

p = (1)(2)(3)(4)(5)(6)(7)(8)(9) . . . (27)(28)(29)(30)
= (1)(2)(3)(4)(5)(2)(3)(7)(8)(3)(3)(10)(11)(3)(4)(13)(14)(3)(5)(16)(17)(3)(3)(2)(19)(20)(3)(7)(22)(23)(3)(8)(25)(26). . . (3)(3)(3)(28)(29)(3)(10)
= (3)^14(other non-3 stuff)

Answer: C

Cheers,
Brent

vikkimba17 wrote:If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

The product of the integers from 1 to 30 inclusive is 30!. Thus, we need to determine the number of factors of 3 in 30!. To do so, we can use the following shortcut in which we divide 30 by 3 then divide the quotient (ignore any nonzero remainder) by 3, and then continue this process until we no longer get a nonzero quotient.

30/3 = 10

10/3 = 3 (we can ignore the remainder)

3/3 = 1

Since 1/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 30!.

Thus, there are 10 + 3 + 1 = 14 factors of 3 within 30!.

Answer: C

Hi vikkimba17,

These types of questions are based on a math concept called "prime factorization", which basically means that any integer greater than 1 is either prime OR the product of a bunch of primes.

Here's a simple example:

24 = (2)(2)(2)(3)

Now, when it comes to this question, we're asked to multiply all the integers from 1 to 30, inclusive and find the greatest integer K for which 3^K is a factor of this really big number.

Here's a simple example with a smaller product:
1 to 6, inclusive...
(1)(2)(3)(4)(5)(6)

Then numbers 1, 2, 4 and 5 do NOT have any 3's in them, so we can essentially ignore them:
3 = one 3
6 = (2)(3) = one 3
Total = two 3's

So 3^2 is the biggest "power of 3" that goes into the product of 1 to 6, inclusive.

Using that same idea, we need to find all of the 3's in the product of 1 to 30, inclusive. Here though, you have to be careful, since there are probably MORE 3's than immediately realize:

3 = one 3
6 = one 3
9 = (3)(3) = two 3s
12 = one 3
15 = one 3
18 = (2)(3)(3) = two 3s
21 = one 3
24 = one 3
27 = (3)(3)(3) = three 3s
30 = one 3

Total = 14 3's

Final Answer: C

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