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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.

Theory that will help you in solving these questions efficiently:

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!?
_________________

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.
_________________

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: \(k=\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\).

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.

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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30 Nov 2012, 12:39

1

This post received 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.

maybe i am way off, but i calculated the answer this way...

i took 20 * 20 = 400, since this represents the largest possible product. then factored 400 to get 2, 2, 2, 2, 5, & 5... whose sum equals 18. thus, k = 18.

seems simple enough, but not sure if this theory holds true.

You are missing the question here: To put it simply, the question is "How many 2s are there in 20!"

20! = 1*2*3*4*5...*19*20 (This is 20 factorial written as 20!)

n = 1*2*3*4*5*6*7.....*19*20

How many 2s are there in n? One 2 from 2 Two 2s from 4 One two from 6 Three 2s from 8 and so on...

Re: if n is the product of integers from 1 to 20 inclusive [#permalink]

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16 Dec 2010, 14:47

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...
_________________

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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30 Nov 2012, 11:58

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?

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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19 Dec 2013, 07:01

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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.

What the question tests is whether in a product you are able to find out how many times multiplication by a certain number happens? In this case it is multiplication by 2.

In the product 1*2*3 upto 20 , multiplication by 2 happens in 2, 4, 6 and in every even number upto 20. So it should be 10 times. However in 4,12 and 20 it happens twice and in 8 it happens thrice and in 16 it happens 4 times. So totally it happens 10 +1+1+1+2+3=18 times.

So we can see the maximum value of K can be 18.
_________________

WE: Business Development (Non-Profit and Government)

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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08 Apr 2014, 10:05

maybe i am way off, but i calculated the answer this way...

i took 20 * 20 = 400, since this represents the largest possible product. then factored 400 to get 2, 2, 2, 2, 5, & 5... whose sum equals 18. thus, k = 18.

seems simple enough, but not sure if this theory holds true.

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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29 Apr 2015, 20:44

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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WE: Project Management (Non-Profit and Government)

Re: If n is the product of integers from 1 to 20 inclusive [#permalink]

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01 Jun 2015, 17:46

VeritasPrepKarishma wrote:

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.

Theory that will help you in solving these questions efficiently:

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!?

Your question is good one.

We have 30! & the number of 2's & 3's are calculated as you told.

30/2 + 30/4 + 30/8 + 30/16 = 15+ 7 + 3 + 1 = 26.

# of 3's are 30/3 + 30/9 + 30/27 = 10 + 3 +1 = 14.

For each 12 we need two 2's & one 3. We have 26 2's so we can make only 13 such pairs.

Thats the catch here the limiting factor is 2 & not the ( greatest prime 3).

In an earlier explanation, you had mentioned to find the maximum power of 6 in 40! factors of 2 & 3 are to be found. And since number of 3's were lesser, we can make 18 6's. Whereas when finding factors of 12 in 30!, 3 yields 14 powers & 2 yields 26 powers. But why the answer is 13 powers instead.?

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