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product of the integers

by gmatblood » Mon Oct 31, 2011 4:31 am
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
Last edited by gmatblood on Mon Oct 31, 2011 8:53 am, edited 1 time in total.
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by pemdas » Mon Oct 31, 2011 4:53 am
del
Last edited by pemdas on Mon Apr 09, 2012 6:45 pm, edited 1 time in total.
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by GMATGuruNY » Mon Oct 31, 2011 5:34 am
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
I believe that 3k is supposed to be 3^k. I've amended the question to reflect its intent.

p = 30!.
3^k = the number of 3's that can be divided into p.
The quickest approach is to count how many times EACH POWER OF 3 can be divided into 30!:

30/3¹ = 10.
The calculation above indicates that 30! includes 10 multiples of 3¹.

30/3² = 30/9 = 3.
The calculation above indicates that 30! includes 3 multiples of 3².

30/3³ = 30/27 = 1.
The calculation above indicates that 30! includes 1 multiple of 3³.

Thus, p includes 10 multiples of 3, 3 multiples of 3², and 1 multiple of 3³.
Thus, the total number of 3's that can be divided into p = 10+3+1 = 14.

The correct answer is C.
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by qazi11 » Thu Dec 29, 2016 1:52 am
This method seems to be working only when factorial of numbers are considered. how about instead of the factorial of a number we have a definite number say 10678 eg
for a number p=10678, what is the greatest integer k for which 2^k is
a factor of p ? .. is there any shortcut for this or it has to be done using prime factorisation of the entire number ?
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by Brent@GMATPrepNow » Thu Dec 29, 2016 8:12 am
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

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by [email protected] » Thu Dec 29, 2016 9:06 am
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.

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by Scott@TargetTestPrep » Wed Jan 04, 2017 7:21 am
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

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by Jay@ManhattanReview » Mon Jan 09, 2017 11:55 pm
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

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