GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 25 Jan 2020, 04:14 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If n is the product of the integers from 1 to 20 inclusive, what is th

Author Message
TAGS:

### Hide Tags

Intern  Joined: 24 Jun 2009
Posts: 34
If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

16 00:00

Difficulty:

(N/A)

Question Stats: 87% (00:59) correct 13% (01:33) wrong based on 1493 sessions

### HideShow 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
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

34
1
23
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

##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

3
2
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: $$\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$

_________________
Director  Joined: 03 Sep 2006
Posts: 602
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

1
1
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 V
Joined: 02 Sep 2009
Posts: 60647
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

14
19
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:
$$k=\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$.

_________________
Manager  Joined: 28 Aug 2010
Posts: 145
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

2
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...
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

4
3
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.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

10
2
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

Intern  Joined: 14 Sep 2012
Posts: 1
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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.
Manager  Joined: 22 Dec 2011
Posts: 207
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

1
1
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 V
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

1
For a detailed discussion on this concept, check out this post: http://www.veritasprep.com/blog/2011/06 ... actorials/
_________________
Karishma
Veritas Prep GMAT Instructor

Director  S
Joined: 17 Dec 2012
Posts: 622
Location: India
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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
Magical Logicians

Holistic and Holy Approach
Intern  Joined: 10 Feb 2014
Posts: 16
Location: United States
Concentration: Strategy, Real Estate
GMAT Date: 06-20-2014
GPA: 3.04
WE: Business Development (Non-Profit and Government)
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

1
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

Manager  B
Joined: 24 May 2014
Posts: 86
Location: India
GMAT 1: 590 Q39 V32
GRE 1: Q159 V151

GRE 2: Q159 V153
GPA: 2.9
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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.?
CEO  S
Joined: 20 Mar 2014
Posts: 2548
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

narendran1990 wrote:
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.?

For maximum power of x (with all prime factors having powers of 1) in n! = integer value of ($$\frac{n}{x}+\frac{n}{x^2}+\frac{n}{x^3}...+\frac{n}{x^k}),$$where $$x^k < n$$.

This for powers of 6 (=2*3) in 30! will be = maximum power of 2 or 3.

Max. power of 2 in 30 ! = 30/2 + 30/(2^2) + 30/(2^3) + 30/ (2^4) = 30/2 + 30/4 + 30/8+30/16 = INTEGER VALUES ONLY = 15+7+3+1 = 26

Max. power of 3 in 30 ! = 30/3 + 30/(3^2) + 30/(3^3) = 30/3 + 30/9 + 30/27 = INTEGER VALUES ONLY = 10+3+1 = 14. So you see 3s are fewer in number than 2s.

Hence the maximum power of 6 in 30! = maximum power of 3 in 30! = 14.

Now coming back to your other question of power of 12 in 30!.

Again 12=$$2^2*3$$, be careful of this now as you need 2 '2s' and 1 '3s' to make 12. Thus, based on calculations done above,

Maximum power of 2 in 30! = 26 ===> maximum ppower of $$2^2$$ = 26/2 = 13.

Maximum power of 3 in 30! = 14.

Thus the more critical case now becoms the maximum power of $$2^2$$ = 13.

Hope this helps.
Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4870
Location: India
GPA: 3.5
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

1
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 the integers from 1 to 20 inclusive = 20!

Highest value of $$2^k$$ is

20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1

Hence the value of k will be 10 + 5 + 2 + 1 = 18

So, The correct answer will be (D) 18
_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Intern  B
Status: preparing for GMAT
Joined: 29 Nov 2017
Posts: 22
GPA: 3.55
WE: Military Officer (Military & Defense)
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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 $$n!$$: http://gmatclub.com/forum/everything-ab ... 85592.html

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

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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 $$n!$$: http://gmatclub.com/forum/everything-ab ... 85592.html

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

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.
_________________
VP  V
Joined: 20 Jul 2017
Posts: 1261
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

### Show Tags

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

Given, n = 20!
Number of multiples of 2 = 10
Number of multiples of 2^2 = 5
Number of multiples of 2^3 = 2
Number of multiples of 2^4 = 1

So, k = 10 + 5 + 2 + 1 = 18

Option D

Posted from my mobile device Re: If n is the product of the integers from 1 to 20 inclusive, what is th   [#permalink] 17 Jul 2019, 05:35

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

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