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If n is a positive integer and 100!/72^n is a positive integer, what

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If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 25 Feb 2020, 01:04
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If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

A. 20
B. 21
C. 22
D. 23
E. 24

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If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 25 Feb 2020, 02:14
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If n is a positive integer and \(\frac{100!}{72^{n}}\) is a positive integer, what is the greatest possible value of \(n\) ?

\(72^{n}= (8*9)^{n}= 2^{3n}*3^{2n}
\)
--> In order to get the greatest possible value of n, finding the number of 3s in 100! is enough:

\([\frac{100}{3}]+[\frac{100}{9}]+[\frac{100}{27}]+[\frac{100}{81}]+...=
\)

=\(33+11+3+1= 48 \) --> the number of \(3s\) in \(100!\)
-->\( 48= 2n\) --> \(n =24 \)

The answer is E.
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Re: If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 25 Feb 2020, 06:35
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Bunuel wrote:
If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

A. 20
B. 21
C. 22
D. 23
E. 24

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72 = 2^3*3^2

So, 72 will have more 2's than 3's and to find the highest power of 72 find the highest power of 3 in 100

100/3 = 33
33/3 = 11
11/3 = 3
3/1 = 1

Total is 33 + 11 + 3 + 1 = 48

Since, power of 3 is 2 we need to divide 48/2 = 24 , Answer must be (E)
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If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 26 Feb 2020, 03:10
Bunuel wrote:
If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

A. 20
B. 21
C. 22
D. 23
E. 24

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Here is the same question with different language.

https://gmatclub.com/forum/f-n-is-defin ... 01844.html

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.....


i.e. Power of prime 3 in 100! = [100/3] + [100/3^2] + [100/3^3] + [100/3^4] + ... and so on

i.e. Power of prime 3 in 100! = 33 + 11 + 3 + 1 + 0 +... and so on = 48

Similarly, Power of 2 in 100! = 50+25+12+6+3+1 = 97

i.e. \(100! = 2^{97}*3^{48}*... = (2^3*3^2)^{24}*2{25}*...= 72^{24}*...\)

I hope it helps!
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If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 28 Feb 2020, 07:43
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Bunuel wrote:
If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

A. 20
B. 21
C. 22
D. 23
E. 24



We are given that 100!/72^n is an integer and need to determine the largest possible integer value of n. Breaking 72n into prime factors, we have:

72^n = 2^3n * 3^2n

We see that since there are fewer 3’s than there are 2’s in 100!, the greatest possible integer value of n is dependent on the number of 3’s in 100!. To determine the number of 3’s, we can use a factorial division shortcut:

100/3 = 33 (ignore the remainder)
100/9 = 11 (ignore the remainder)
100/27 = 3 (ignore the remainder)
100/81 = 1 (ignore the remainder)
100/243 = 0

Since 100/243 has a quotient of zero, we can stop.

We see that there are 33 + 11 + 3 + 1 = 48 threes in 30!.

However, since we need to determine 3^2n, then the largest value of n such that 100!/3^2n is an integer is 48/2 = 24.

Answer: E
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Re: If n is a positive integer and 100!/72^n is a positive integer, what  [#permalink]

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New post 28 Mar 2020, 03:32
Bunuel wrote:
If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

A. 20
B. 21
C. 22
D. 23
E. 24

Project PS Butler


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Asked: If n is a positive integer and \(\frac{100!}{72^n}\) is a positive integer, what is the greatest possible value of n ?

72 = 2^3*3^2
Power of 2 in 100! = 50 + 25 + 12 + 6 + 3 + 1 = 97
Power of 2^3 in 100! = 32
Power of 3 in 100! = 33 + 11 + 3 + 1 = 48
Power of 3^2 in 100! = 24

Power of 72 in 100! = min (32,24) = 24

IMO E
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Re: If n is a positive integer and 100!/72^n is a positive integer, what   [#permalink] 28 Mar 2020, 03:32

If n is a positive integer and 100!/72^n is a positive integer, what

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