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# If n is the product of integers from 1 to 20 inclusive

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If n is the product of integers from 1 to 20 inclusive [#permalink]  14 Dec 2010, 17:31
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85% (01:34) correct 14% (00:57) wrong based on 259 sessions
If n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2^k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

any efficient way to solve such questions.
[Reveal] Spoiler: OA

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Ajit

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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  14 Dec 2010, 19:23
5
KUDOS
Expert's post
ajit257 wrote:
Q. if n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2^k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

any efficient way to solve such questions.

I will take a simpler example first.

What is the greatest value of k such that 2^k is a factor of 10! ?
We need to find the number of 2s in 10!
Method:
Step 1: 10/2 = 5
Step 2: 5/2 = 2
Step 3: 2/2 = 1
Step 4: Add all: 5 + 2 + 1 = 8 (Answer)

Logic:
10! = 1*2*3*4*5*6*7*8*9*10
Every alternate number will have a 2. Out of 10 numbers, 5 numbers will have a 2. (Hence Step 1: 10/2 = 5)
These 5 numbers are 2, 4, 6, 8, 10
Now out of these 5 numbers, every alternate number will have another 2 since it will be a multiple of 4 (Hence Step 2: 5/2 = 2)
These 2 numbers will be 4 and 8.
Out of these 2 numbers, every alternate number will have yet another 2 because it will be a multiple of 8. (Hence Step 3: 2/2 = 1)
This single number is 8.

Now all 2s are accounted for. Just add them 5 + 2 + 1 = 8 (Hence Step 4)
These are the number of 2s in 10!.

Similarly, you can find maximum power of any prime number in any factorial.
If the question says 4^m, then just find the number of 2s and half it.
If the question says 6^m, then find the number of 3s and that will be your answer (because to make a 6, you need a 3 and a 2. You have definitely more 2s in 10! than 3s. So number of 3s is your limiting condition.)
Let's take this example: Maximum power of 6 in 40!.
40/3 = 13
13/3 = 4
4/3 = 1
Total number of 3s = 13 + 4 + 1 = 18
40/2 = 20
20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38
Definitely, number of 3s are less so we can make only 18 6s in spite of having many more 2s.
Usually, the greatest prime number will be the limiting condition.

Perhaps you can answer your question yourself now.... and also answer one of mine: What happens if I ask for the greatest power of 12 in 30!?
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3758 Location: Pune, India Followers: 806 Kudos [?]: 3205 [3] , given: 137 Re: if n is the product of integers from 1 to 20 inclusive [#permalink] 16 Dec 2010, 19:06 3 This post received KUDOS Expert's post ajit257 wrote: Hi Karishma, Thanks a lot for the concept. So to answer your question we have (10 + 9+ 3+1 = 23 ...3s) and (15 + 7 + 3 + 1 = 26 ...2s) so we should get 23 12s. Thanks a lot... 30/3 = 10 10/3 = 3 3/3 = 1 Total number of 3s = 14 30/2 = 15 15/2 = 7 7/2 = 3 3/2 = 1 Total number of 2s = 26 But to make a 12, we need two 2s and one 3. Hence, out of 26 2s, we can make only 13 12's. Therefore, the maximum power of 12 in 30! is 13. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  15 Dec 2010, 01:07
2
KUDOS
Expert's post
ajit257 wrote:
Q. if n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2^k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

any efficient way to solve such questions.

Check this: everything-about-factorials-on-the-gmat-85592.html other examples: facorial-ps-105746.html#p827453

Finding the highest powers of a prime number k, in the n!

What is the power of 3 in 35!?

The formula is:
\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3} ... till n>k^x

For example what is the highest power of 3 in 35!:
\frac{35}{3}+\frac{35}{9}+\frac{35}{27}=11+3+1=15, so the highest power of 3 in 35! is 15: 3^{15}*x=35!, where x is the product of all other factors of 35!.

Back to the original question:
If n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2^k is a factor of n?
A. 10
B. 12
C. 15
D. 18
E. 20

Given: n=20!. The highest power k for which 2^k is a factor of n can be found with the above formula:
k=\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18.

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Re: If n is the product of integers from 1 to 20 inclusive [#permalink]  30 Nov 2012, 11:39
1
KUDOS
aman1988 wrote:
Hi,

Thanks to everyone for the explanation.

I just have one question that will render my doubt on this topic clear.

When you discussed about "How many 12s in 40!", I got till the point that there are 14 3s and 26 2s and that 2^2 *3 makes one 12.
But what I didn't understand is after this step, how did we reach to the conclusion that there will be 13 12s? Why not 14 12s?

Thanks a lot.
Aman.

What Bunuel and Karishma mean is that to form 12 we need one pair of 2s and one 3
so from twenty six 2s how many pairs of 2s can be formed exactly 13 .. and each of these pair will need a 3 in it to make each of these 12. so 13 3s are used. One 3 is left over with out any pair.

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Re: If n is the product of integers from 1 to 20 inclusive [#permalink]  06 Dec 2012, 19:28
1
KUDOS
Expert's post
aman1988 wrote:
oops.
I just realized this is a very old thread.

For a detailed discussion on this concept, check out this post: http://www.veritasprep.com/blog/2011/06 ... actorials/
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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  14 Dec 2010, 18:31
n = 20!

I could not find any fast method, but just checked the number of factors of 2 ( all the even terms will have ), and it turns out to be

2^(18)

and k=18.

20*18*16*14*12*10*8*6*4*2
(5*2*2)*(9*2)*(2*2*2*2)*(3*2*2)*(5*2)*(2*2*2)*(3*2)*(2*2)*(2)

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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  16 Dec 2010, 13:47
Hi Karishma,
Thanks a lot for the concept. So to answer your question we have (10 + 9+ 3+1 = 23 ...3s) and (15 + 7 + 3 + 1 = 26 ...2s) so we should get 23 12s. Thanks a lot...
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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  16 Dec 2010, 14:05
Expert's post
ajit257 wrote:
Hi Karishma,
Thanks a lot for the concept. So to answer your question we have (10 + 9+ 3+1 = 23 ...3s) and (15 + 7 + 3 + 1 = 26 ...2s) so we should get 23 12s. Thanks a lot...

Question: what is the highest power of 12=2^2*3 in 30!?

Now, you are right saying that the highest power of 2 in 30! is 26 as \frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26 but the highest power of 3 in 30! is 14 (not 23) as \frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14. Next, as 12=2^2*3 you'll need twice as many 2-s as 3-s so 26 2-s is enough for 13 3-s, which means that the highest power of 12 in 30! is 13. Or in another way: we got that 30!=2^{26}*3^{14}*k, where k is th product of all other multiples of 30! (other than 2 and 3) --> 30!=2^{26}*3^{14}*k=(2^2*3)^{13}*3*k=12^{13}*3*k.

Check the links in my previous post for more examples.

Hope it's clear.
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Re: if n is the product of integers from 1 to 20 inclusive [#permalink]  16 Dec 2010, 17:04
oh yes ...calc mistake......thanks a ton Bunuel. It is indeed 13 3s
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Re: If n is the product of integers from 1 to 20 inclusive [#permalink]  30 Nov 2012, 10:58
Hi,

Thanks to everyone for the explanation.

I just have one question that will render my doubt on this topic clear.

When you discussed about "How many 12s in 40!", I got till the point that there are 14 3s and 26 2s and that 2^2 *3 makes one 12.
But what I didn't understand is after this step, how did we reach to the conclusion that there will be 13 12s? Why not 14 12s?

Thanks a lot.
Aman.
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Re: If n is the product of integers from 1 to 20 inclusive [#permalink]  30 Nov 2012, 11:16
oops.
I just realized this is a very old thread.
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Re: If n is the product of integers from 1 to 20 inclusive [#permalink]  30 Nov 2012, 23:57
Ok, I got it now. Thank you Sir.
Re: If n is the product of integers from 1 to 20 inclusive   [#permalink] 30 Nov 2012, 23:57
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