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
Number properties
This topic has expert replies
- DavidG@VeritasPrep
- Legendary Member
- Posts: 2663
- Joined: Wed Jan 14, 2015 8:25 am
- Location: Boston, MA
- Thanked: 1153 times
- Followed by:128 members
- GMAT Score:770
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.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
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
-
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Mon Aug 04, 2014 5:39 pm
- Thanked: 1 times
- Followed by:2 members
Hi DavidGDavidG@VeritasPrep wrote: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.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
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
Thanks a lot for the explanation!
Makes perfect sense now
GMAT/MBA Expert
- ceilidh.erickson
- GMAT Instructor
- Posts: 2095
- Joined: Tue Dec 04, 2012 3:22 pm
- Thanked: 1443 times
- Followed by:247 members
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:
https://www.beatthegmat.com/tough-multip ... tml#715830
Btw, what is the source of this question?
And here's another 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
https://www.beatthegmat.com/tough-multip ... tml#715830
Btw, what is the source of this question?
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
- DavidG@VeritasPrep
- Legendary Member
- Posts: 2663
- Joined: Wed Jan 14, 2015 8:25 am
- Location: Boston, MA
- Thanked: 1153 times
- Followed by:128 members
- GMAT Score:770
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.htmlHi DavidG
Thanks a lot for the explanation!
Makes perfect sense now Smile
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
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.
"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.
-
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Mon Aug 04, 2014 5:39 pm
- Thanked: 1 times
- Followed by:2 members
Hi Ceilidhceilidh.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:
And here's another 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
https://www.beatthegmat.com/tough-multip ... tml#715830
Btw, what is the source of this question?
Thank you for the extra questions - highly appreciated!
I got the in a GMAT-test through the Veritas homepage.
-
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Mon Aug 04, 2014 5:39 pm
- Thanked: 1 times
- Followed by:2 members
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
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