check how many 6's available in 25!
6(2*3),6,12,18,24=5 6's
Pick A
If 6^m divides 25!, then...
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selango
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Sorry.Let me explain in detail.
Given 6^m divides 25!
25!/6^m
We need to find number of 6's occurs in 25! so that we can cancel 6's occuring in both numerator and denominator.
Now how many 6's are in 25!.
6=2*3
6=1*6
12=2*6
18=3*6
24=4*6
As you can see there are five 6's in 25!.Now if we sub m=6.
25!/6^6
Now 6's occurring in both numerator and denominator cancels.
So m=6.
Hope this clarify!!
Given 6^m divides 25!
25!/6^m
We need to find number of 6's occurs in 25! so that we can cancel 6's occuring in both numerator and denominator.
Now how many 6's are in 25!.
6=2*3
6=1*6
12=2*6
18=3*6
24=4*6
As you can see there are five 6's in 25!.Now if we sub m=6.
25!/6^6
Now 6's occurring in both numerator and denominator cancels.
So m=6.
Hope this clarify!!
--Anand--
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nxanand
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You have to find out how many 2's and 3's you have in 25!. That will give you the maximum number of 6's you can have in 25!.
represented below all the numbers from 1-25 that has atleast a '2' or a '3'.
2, 3, 2^2, 2*3, 2^3, 3^2, 2*5, 2^2*3, 2*7, 3*5, 2^4, 2*3^2, 2^2*5, 3*7, 2*11, 2^3*3
this gives 2^22 & 3^10
so the maximum no.of 6s possible is 10
therefore answer is C.
represented below all the numbers from 1-25 that has atleast a '2' or a '3'.
2, 3, 2^2, 2*3, 2^3, 3^2, 2*5, 2^2*3, 2*7, 3*5, 2^4, 2*3^2, 2^2*5, 3*7, 2*11, 2^3*3
this gives 2^22 & 3^10
so the maximum no.of 6s possible is 10
therefore answer is C.
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euro
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I tried to solve this in MS-Excel and found that 6^14 evenly divides 25!.euro wrote:Can somebody help with this one....
(1) If 6^m divides 25! then what is the largest value of m?
(A) 5
(B) 8
(C) 10
(D) 22
(E) 26
However, 14 is not there in the options. From the given alternatives, 10 appears to be the answer based on nxanand's explanation.
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Brian@VeritasPrep
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Great question here - but I have a few things to point out toward the correct answer:
1) Beware of the 'too-easy' choice A here...26 (choice E) is a long, long way from 5, so that should tip you off that maybe you're missing something.
2) What the initial explanation misses is that we don't simply need "naturally occurring" 6s - 6, 12, 18, etc. We can "manufacture" 6s out of any combination of 2 and 3.
Consider this - is 3*4 divisible by 6? It is - 3*4 is 12, which divides by 6. It just doesn't have a "natural" 6 in it, but if we break it down to prime factors it's 3*2*2, and in order to be divisible by 6 we only need a 2 and a 3.
So...for problems like these, break the divisor down to prime factors and find as many pairs of that factorization as you can.
Here, we need 2*3. Strategically, there will be many more 2s than 3s - every SECOND number gives us a 2, but every THIRD number gives us a 3. 3s are going to be our "constraint" then, so all we really need to do is find the number of 3s in 25!. Looking at that, we have as muttiples of 3:
3, 6, 9, 12, 15, 18, 21, 24
But in keeping in step with the "prime factors" consideration - any 3 will do - we need to break these out to maximize our 3s. 9 = 3*3, for example, so it gives us two 3s, not just one. From our list we have:
3
3*2
3*3
3*4
3*5
3*3*2
3*7
3*8
Because each of 9 and 18 gives us an extra 3, we have a total of ten 3s, so the correct answer is 10.
A few big takeaways here:
1) With divisibility-related questions, it's usually helpful to break numbers down to prime factors.
2) Use your judgment with this one, but if a question almost seems too easy and you can't justify any other answer choices as being remotely possible (here if you picked 5 the others seem way, way far away with no basis for picking them) you probably forgot to consider something and you may want to return to the question more thoroughly.
1) Beware of the 'too-easy' choice A here...26 (choice E) is a long, long way from 5, so that should tip you off that maybe you're missing something.
2) What the initial explanation misses is that we don't simply need "naturally occurring" 6s - 6, 12, 18, etc. We can "manufacture" 6s out of any combination of 2 and 3.
Consider this - is 3*4 divisible by 6? It is - 3*4 is 12, which divides by 6. It just doesn't have a "natural" 6 in it, but if we break it down to prime factors it's 3*2*2, and in order to be divisible by 6 we only need a 2 and a 3.
So...for problems like these, break the divisor down to prime factors and find as many pairs of that factorization as you can.
Here, we need 2*3. Strategically, there will be many more 2s than 3s - every SECOND number gives us a 2, but every THIRD number gives us a 3. 3s are going to be our "constraint" then, so all we really need to do is find the number of 3s in 25!. Looking at that, we have as muttiples of 3:
3, 6, 9, 12, 15, 18, 21, 24
But in keeping in step with the "prime factors" consideration - any 3 will do - we need to break these out to maximize our 3s. 9 = 3*3, for example, so it gives us two 3s, not just one. From our list we have:
3
3*2
3*3
3*4
3*5
3*3*2
3*7
3*8
Because each of 9 and 18 gives us an extra 3, we have a total of ten 3s, so the correct answer is 10.
A few big takeaways here:
1) With divisibility-related questions, it's usually helpful to break numbers down to prime factors.
2) Use your judgment with this one, but if a question almost seems too easy and you can't justify any other answer choices as being remotely possible (here if you picked 5 the others seem way, way far away with no basis for picking them) you probably forgot to consider something and you may want to return to the question more thoroughly.
Last edited by Brian@VeritasPrep on Tue Sep 14, 2010 11:04 am, edited 1 time in total.
Brian Galvin
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Brian@VeritasPrep
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Hey Euro,
Good point on using Excel to check your work on these - funny story...it was a problem just like this for which I needed to write a solution for our materials a few years ago, and my boss checked my work using Excel. I still remember him walking over with a look of sheer disappointment on his face that I was "so far off". But as it turns out, Excel rounds off huge numbers like 25! so that it won't actually give you a precise answer to the decimal point, mainly because there's no reason that it would ever have to. 25! is a "NASA number" - unless you're dealing with huge distances in space or the number of bacteria in a lake or something like that, you'll never use 25!, and if you're dealing with a huge number like that precision to the decimal place doesn't matter...it's really insignificant. Excel will round those numbers at a certain point - which is why the GMAT loves them. NASA doesn't care to calculate them, but by using your problem-solving skills you can determine the right answer!
Good point on using Excel to check your work on these - funny story...it was a problem just like this for which I needed to write a solution for our materials a few years ago, and my boss checked my work using Excel. I still remember him walking over with a look of sheer disappointment on his face that I was "so far off". But as it turns out, Excel rounds off huge numbers like 25! so that it won't actually give you a precise answer to the decimal point, mainly because there's no reason that it would ever have to. 25! is a "NASA number" - unless you're dealing with huge distances in space or the number of bacteria in a lake or something like that, you'll never use 25!, and if you're dealing with a huge number like that precision to the decimal place doesn't matter...it's really insignificant. Excel will round those numbers at a certain point - which is why the GMAT loves them. NASA doesn't care to calculate them, but by using your problem-solving skills you can determine the right answer!
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
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