If n is the product of integers from 1 to 20 inclusive : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 17:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 28 Aug 2010
Posts: 260
Followers: 6

Kudos [?]: 552 [2] , given: 11

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

### Show Tags

14 Dec 2010, 17:31
2
KUDOS
22
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

85% (01:41) correct 15% (01:05) wrong based on 1293 sessions

### HideShow timer Statistics

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

_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7121
Location: Pune, India
Followers: 2133

Kudos [?]: 13640 [24] , given: 222

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

### Show Tags

14 Dec 2010, 19:23
24
KUDOS
Expert's post
11
This post was
BOOKMARKED
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!?
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13640 [8] , given: 222 Re: if n is the product of integers from 1 to 20 inclusive [#permalink] ### Show Tags 16 Dec 2010, 19:06 8 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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 Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93114 [6] , given: 10552

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

### Show Tags

15 Dec 2010, 01:07
6
KUDOS
Expert's post
13
This post was
BOOKMARKED
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:
$$k=\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$.

_________________
Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 769 [1] , given: 33

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

### Show Tags

14 Dec 2010, 18:31
1
KUDOS
1
This post was
BOOKMARKED
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.

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

Leads to$$2^(18)$$
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077

Kudos [?]: 93114 [1] , given: 10552

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

### Show Tags

16 Dec 2010, 14:05
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.
_________________
Senior Manager
Joined: 22 Dec 2011
Posts: 298
Followers: 3

Kudos [?]: 233 [1] , given: 32

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

### Show Tags

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.

Cheers
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7121
Location: Pune, India
Followers: 2133

Kudos [?]: 13640 [1] , given: 222

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

### Show Tags

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/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13640 [1] , given: 222 Re: If n is the product of integers from 1 to 20 inclusive [#permalink] ### Show Tags 08 Apr 2014, 21:34 1 This post received KUDOS Expert's post jmyer028 wrote: 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... When you count them all, you get 18. But there are more efficient ways of doing this discussed in this post: http://www.veritasprep.com/blog/2011/06 ... actorials/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Senior Manager
Joined: 28 Aug 2010
Posts: 260
Followers: 6

Kudos [?]: 552 [0], given: 11

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

### Show Tags

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...
_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Senior Manager
Joined: 28 Aug 2010
Posts: 260
Followers: 6

Kudos [?]: 552 [0], given: 11

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

### Show Tags

16 Dec 2010, 17:04
oh yes ...calc mistake......thanks a ton Bunuel. It is indeed 13 3s
_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Intern
Joined: 14 Sep 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 4

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

### Show Tags

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.
Intern
Joined: 14 Sep 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 4

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

### Show Tags

30 Nov 2012, 11:16
oops.
I just realized this is a very old thread.
Intern
Joined: 14 Sep 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 4

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

### Show Tags

30 Nov 2012, 23:57
Ok, I got it now. Thank you Sir.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13440
Followers: 575

Kudos [?]: 163 [0], given: 0

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

### Show Tags

19 Dec 2013, 06:01
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.
_________________
Senior Manager
Joined: 17 Dec 2012
Posts: 447
Location: India
Followers: 26

Kudos [?]: 395 [0], given: 14

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

### Show Tags

05 Apr 2014, 22:00
ajit257 wrote:
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.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom and Online Coaching

Intern
Joined: 10 Feb 2014
Posts: 17
Location: United States
Concentration: Strategy, Real Estate
GMAT Date: 06-20-2014
GPA: 3.04
WE: Business Development (Non-Profit and Government)
Followers: 0

Kudos [?]: 10 [0], given: 0

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

### Show Tags

08 Apr 2014, 09:05
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.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13440
Followers: 575

Kudos [?]: 163 [0], given: 0

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

### Show Tags

29 Apr 2015, 19:44
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.
_________________
Manager
Joined: 21 Jul 2014
Posts: 71
Location: United States
WE: Project Management (Non-Profit and Government)
Followers: 0

Kudos [?]: 11 [0], given: 58

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

### Show Tags

01 Jun 2015, 16:46
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.

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!?

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).

So we will have 13 such pairs of two 2's & one 3.

let me know if the answers correct.
Manager
Joined: 24 May 2014
Posts: 99
Location: India
GMAT 1: 590 Q39 V32
GRE 1: 310 Q159 V151
GRE 2: 312 Q159 V153
GPA: 2.9
Followers: 0

Kudos [?]: 7 [0], given: 348

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

### Show Tags

30 Nov 2015, 06:01
VeritasPrepKarishma

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.?
Re: If n is the product of integers from 1 to 20 inclusive   [#permalink] 30 Nov 2015, 06:01

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
3 If n is the product of all the integers from 5 to 20, inclusive, what 4 11 Nov 2016, 00:26
11 If n is the product of the integers from 1 to 8, inclusive 13 09 Jul 2012, 02:31
3 If n is the product of the integers from 1 to 20 inclusive, 6 22 Nov 2009, 17:05
29 If p is the product of the integers from 1 to 30, inclusive, 11 16 Sep 2009, 09:54
1 If n is the product of the integers from 1 to 8, inclusive, 4 22 Aug 2007, 10:22
Display posts from previous: Sort by