NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 02:38 It is currently 25 Apr 2024, 02:38
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If n is the product of the integers from 1 to 20 inclusive, what is th

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
shanewyatt
Intern
Intern
Joined: 24 Jun 2009
Posts: 31
Own Kudos [?]: 639 [78]
Given Kudos: 2
Send PM
If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   22 Nov 2009, 18:05
3
Kudos
75
Bookmarks
00:00
A
B
C
D
E

Difficulty:

5% (low)

Question Stats:

86% (00:59) correct 14% (01:31) wrong based on 2709 sessions
Hide Show timer Statistics
If n is the product of the 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
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64907 [93]
Given Kudos: 426
Location: Pune, India
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   14 Dec 2010, 20:23
49
Kudos
43
Bookmarks
Expert Reply
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!?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618887 [35]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   15 Dec 2010, 02:07
14
Kudos
21
Bookmarks
Expert Reply
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:
\(k=\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\).

Answer: D.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618887 [21]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   22 Nov 2009, 18:25
10
Kudos
10
Bookmarks
Expert Reply
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): everything-about-factorials-on-the-gmat-85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.
General Discussion
User avatar
LM
Senior Manager
Senior Manager
Joined: 03 Sep 2006
Posts: 446
Own Kudos [?]: 6776 [4]
Given Kudos: 33
Send PM
GMAT ToolKit User
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   14 Dec 2010, 19:31
3
Kudos
1
Bookmarks
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\(2^(18)\)
User avatar
ajit257
Manager
Manager
Joined: 28 Aug 2010
Posts: 124
Own Kudos [?]: 3525 [2]
Given Kudos: 11
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   16 Dec 2010, 14:47
2
Kudos
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...
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618887 [10]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   16 Dec 2010, 15:05
5
Kudos
4
Bookmarks
Expert Reply
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.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64907 [14]
Given Kudos: 426
Location: Pune, India
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   16 Dec 2010, 20:06
12
Kudos
2
Bookmarks
Expert Reply
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.
avatar
aman1988
Intern
Intern
Joined: 14 Sep 2012
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 4
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   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?

Thanks a lot.
Aman.
User avatar
Jp27
Manager
Manager
Joined: 22 Dec 2011
Posts: 175
Own Kudos [?]: 1042 [2]
Given Kudos: 32
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   30 Nov 2012, 12:39
1
Kudos
1
Bookmarks
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
User avatar
S
SVaidyaraman
Director
Director
Joined: 17 Dec 2012
Posts: 589
Own Kudos [?]: 1519 [0]
Given Kudos: 20
Location: India
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   05 Apr 2014, 23:00
Expert Reply
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.
avatar
jmyer028
Intern
Intern
Joined: 10 Feb 2014
Posts: 13
Own Kudos [?]: 26 [0]
Given Kudos: 0
Location: United States
Concentration: Strategy, Real Estate
Schools: Duke '17 Darden '17 Kenan-Flagler '17 McDonough '17
GMAT Date: 06-20-2014
GPA: 3.04
WE:Business Development (Non-Profit and Government)
Send PM
GMAT ToolKit User
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   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.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64907 [2]
Given Kudos: 426
Location: Pune, India
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   08 Apr 2014, 22:34
2
Kudos
Expert Reply
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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... actorials/
User avatar
B
destinyawaits
Intern
Intern
Joined: 29 Nov 2017
Status:preparing for GMAT
Posts: 19
Own Kudos [?]: 22 [0]
Given Kudos: 3
GPA: 3.55
WE:Military Officer (Military & Defense)
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   16 Jan 2018, 10:37
Bunuel wrote:
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): https://gmatclub.com/forum/everything-ab ... 85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.


Hey bunuel, this is a very useful formula in finding the highest power for which 2 is a factor of n.

Could you share some possible variations of this? how could it become more difficult?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618887 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   16 Jan 2018, 10:46
Expert Reply
destinyawaits wrote:
Bunuel wrote:
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): https://gmatclub.com/forum/everything-ab ... 85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.


Hey bunuel, this is a very useful formula in finding the highest power for which 2 is a factor of n.

Could you share some possible variations of this? how could it become more difficult?


Everything about Factorials on the GMAT
Power of a Number in a Factorial Problems
Trailing Zeros Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
User avatar
L
GMATinsight
GMAT Club Legend
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 5957
Own Kudos [?]: 13387 [0]
Given Kudos: 124
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Send PM
Reviews Badge
If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   24 Jan 2020, 22:59
Expert Reply
shanewyatt wrote:
If n is the product of the 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


Product of all integers from 1 to 20 = 1*2*3*---*20 = 20!

Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)
i.e. [100/3] = [33.33] = 33
i.e. [100/9] = [11.11] = 11 etc.
Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....


So, Power of 2 in 20! = [20/2] + [20/2^2] + [20/2^3] + [20/2^4] = 10+5+2+1 = 18

Answer: Option D
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64907 [1]
Given Kudos: 426
Location: Pune, India
Send PM
If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   09 Jun 2022, 21:47
1
Kudos
Expert Reply
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.


aman1988

Think about it with a simpler example. If we have five 2s and four 3s, how many 6s can we make?
To make a 6, we need one 2 and one 3. Can we make do with just two 2s? or just two 3s? No. They will give us 4 and 9 respectively.

Now, if we have five 2s but only four 3s, we can make only four 6s (because 3s are the limiting quantity)

Similarly, when we have 14 3s and 26 2s, to make a 12, we need one 3 and a pair of 2s. With 26 2s, we have 13 pairs of 2s.
So we have 14 3s but only 13 pairs of 2s. Hence we can make only 13 12s.
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32665
Own Kudos [?]: 821 [0]
Given Kudos: 0
Send PM
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink] New post   05 Feb 2024, 01:36
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.
GMAT Club Bot
gmatclubot
Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
05 Feb 2024, 01:36
Moderators:
Bunuel
Math Expert
92912 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions