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Product of 1 to 30

Expert replies

by 800_or_bust » Fri Jul 01, 2016 8:15 am
kumar720 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

OA[spoiler]:C[/spoiler]

Help me solving the problem
Just count all the factors of 3 in 30!.

3 x 3 x 3^2 x 3 x 3 x 3^2 x 3 x 3 x 3^3 x 3 = 3^14. Each multiple of 3 in the product will add one additional factor of 3, each multiple of 9 will add one more on top of that, and each multiple of 27 will add one more on top of that.

From left to right, the 3s come from the following integers in the product: 3,6,9,12,15,18,21,24,27,30. You can see 9 and 18 have a total of 2 factors of 3, and 27 has 3.

No need to memorize any sort of formula to answer this... It should take no more than a minute to count the factors of 3.
800 or bust!
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by Brent@GMATPrepNow » Fri Jul 01, 2016 9:50 am
kumar720 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

OA[spoiler]:C[/spoiler]
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
Brent Hanneson - Creator of GMATPrepNow.com
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