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If p is the product of the integers from 1 to 30,

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by Jay@ManhattanReview » Fri Mar 24, 2017 12:12 am
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
Hi vikkimba17,

Pl. post the correct question. It's not '3k,' it's '3^k.' Moreover, do mention the correct answer and the source of the question.

We have to find the greatest value of k such that 3^k is a factor of 30! (1*2*3...*30).

Thus, we have to count the total number of 3s in 30!.

It would be given by quotients of [30/3 + 30/(3^2) + 30/(3^3) + ......]; we must not add '30/(3^4)' since 3^4 = 81 > 30.

Thus the greatest value of k = quotients of [30/3 + 30/(3^2) + 30/(3^3)] = quotients of [30/3 + 30/9 + 30/27] = 10 + 3 + 1 = 14.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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by DavidG@VeritasPrep » Fri Mar 24, 2017 4:14 am
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 method Jay described is the most efficient way to solve this problem, but because the number of multiples of 3 is reasonably small, it's also feasible to employ brute force, and just list them out and count the number of threes in each number's prime factorization.

3 --> 1 three
6 --> 1 three
9 --> 2 threes (9 = 3^2)
12--> 1 three
15 --> 1 three
18 --> 2 threes (18 = 2 * 3^2)
21---> 1 three
24 -->1 three
27 --> 3 three's (27 = 3^3)
30 ---> 1 three

1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 3+ 1 = 14; The answer is C
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by DavidG@VeritasPrep » Fri Mar 24, 2017 4:21 am
For a lengthier post describing the mechanics of such problems, see here: https://www.beatthegmat.com/mba/2011/07/ ... r-problems
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by Brent@GMATPrepNow » Fri Mar 24, 2017 4:40 am
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
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by GMATGuruNY » Fri Mar 24, 2017 5:08 am
I posted a solution here:
https://www.beatthegmat.com/product-of-t ... 94247.html
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by Scott@TargetTestPrep » Thu Mar 30, 2017 3:36 pm
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

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by [email protected] » Thu Mar 30, 2017 5:26 pm
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

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