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

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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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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: $$\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$

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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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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$$2^(18)$$
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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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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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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.
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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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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
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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.
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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.
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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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/
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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: $$\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$

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?
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th [#permalink]
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: $$\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18$$

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

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