lucas211 wrote:if 6^y is a factor of (10!)^2, what is the greatest possible value of y?

a) 2
b) 4
c) 6
d) 8
e) 10

Thanks in advance

If 6^y is a factor of (10!)^2, it means that (10!)^2/(6^y) is an integer, and that (10!)^2/[3^y * 2^y] is an integer.

Put another way, the upper limit for our value of y will be dependent on how many pairs of 2's and 3's are in (10!)^2. Because there will be far more 2's than 3's, y's upper limit will be determined by how many 3's are in (10!)^2.

The multiples of 3 in 10! are 3, 6, and 9.

3---> contains one 3
6 ---> 2*3 ----> contains one 3
9----> 3^2 ----> contains two 3's

So 10! contains a total of four 3's.

Thus (10!)^2 will contain a total of eight 3's.

Answer is D

DavidG@VeritasPrep wrote:

lucas211 wrote:if 6^y is a factor of (10!)^2, what is the greatest possible value of y?

a) 2
b) 4
c) 6
d) 8
e) 10

Thanks in advance

If 6^y is a factor of (10!)^2, it means that (10!)^2/(6^y) is an integer, and that (10!)^2/[3^y * 2^y] is an integer.

Put another way, the upper limit for our value of y will be dependent on how many pairs of 2's and 3's are in (10!)^2. Because there will be far more 2's than 3's, y's upper limit will be determined by how many 3's are in (10!)^2.

The multiples of 3 in 10! are 3, 6, and 9.

3---> contains one 3
6 ---> 2*3 ----> contains one 3
9----> 3^2 ----> contains two 3's

So 10! contains a total of four 3's.

Thus (10!)^2 will contain a total of eight 3's.

Answer is D

Hi DavidG

Thanks a lot for the explanation!
Makes perfect sense now

Hi DavidG

Thanks a lot for the explanation!
Makes perfect sense now Smile

And if you really just cannot get enough of this type of question, here's yet another: https://www.beatthegmat.com/if-x-is-the- ... 70022.html

Another way of think of these equations is phrasing them this way:

"How many 6s can be found in the factorization of 10! * 10! ?"

and if the number sought is NOT prime (such as 6), you can reduce the problem to finding the number of the least common prime in that number. For instance, 6 = 3 * 2. There will be more 2s than 3s in 10! * 10!, so we really only need to count the number of 3s: for every 3 we find, there will always be a 2 we can find to go with it.

With that in mind, the question becomes

"How many 3s can be found in the prime factorization of 10! * 10! ?"

and we can count them as follows:

10! * 10! = 10 * 9 * 8 * 7 * 6 * 5 * 3 * 2 * 1 * 10 * 9 * 8 * 7 * 6 * 5 * 3 * 2 * 1

That means we have 9 * 9 * 6 * 6 * 3 * 3, or 3*3 * 3*3 * 2*3 * 2*3 * 3 * 3, for a total of EIGHT 3s.

ceilidh.erickson wrote:This question is a variation of PS #140 in OG 2016. Here is the full text of that question if you want to practice on a similar one:

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

And here's another similar one:
https://www.beatthegmat.com/tough-multip ... tml#715830

Btw, what is the source of this question?

Hi Ceilidh

Thank you for the extra questions - highly appreciated!
I got the in a GMAT-test through the Veritas homepage.

Hello everybody

Thanks for all your replies!

After I have tried the additional questions you all have given me here, it seems to me that we have a pattern in questions like this, in which we
can divide the number by the least common prime factor. The result (not including decimals) can then be divided with the least common prime factor again, and so on, until we reach "0".

As an illustration with my original question: (if 6^y is a factor of (10!)^2, what is the greatest possible value of y?)

(10!^2) / 6^y

10!^2 / (2^y * 3^y) = so the least common prime factor is 3.

following the above mentioned:
10/3 = 3 (excluding some decimals)
3/3 = 1 or a total of "4"

and since we have 10! squared we have to multiply by 2 for a total of "8".

is this just an incident or would this be a valid approach on testday?

Hope the above makes sense, as I am not familiar with all the mathematical expressions

Thanks in advance