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

Hi qazi11,

If you're given a single 'big' number to work with, then prime-factorization is likely going to be the best way to break that number down into it's 'pieces.'

It's also worth noting that NOTHING about the design of a GMAT question is 'random' - both the wording and the numbers are specifically chosen. Here, with 10678, we can immediately see that 2 is divisible...but there isn't anything else that's obviously a factor (not 3, 5, 9, any other even number, etc.) - which is meant to say that this is not a number that the GMAT would likely ask you to prime factor.

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

gmatblood 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

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 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

gmatblood 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

IMO: 18

Basically, we have to find out the number of 3s in the product of 1 to 30, inclusive.

All the multiples of 3 from 1 to 30 would have at least one 3.

The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, and 30. There are 10 numbers of 3s.

A few multiple of 3 would render two 3s: 9, 18, and 27. So let's count them twice. So, total 3s so far = 10 + 3 = 13.

A few multiple of 3 would render three 3s: 27. So let's count them thrice. So, total 3s so far = 10 + 3 + 1 = 14.

We need not seek out for multiples of 3^4 as 3^4 = 81 > 30.

Thus, the answer = 14. C

-Jay
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