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What is the greatest positive integer x such that 24^x is a factor of

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What is the greatest positive integer x such that 24^x is a factor of  [#permalink]

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New post Updated on: 11 Feb 2020, 23:10
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What is the greatest positive integer x such that 24^x is a factor of 100!?​

A. 32​
B. 33​
C. 46​
D. 47​
E. 48​

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Originally posted by kapil1 on 11 Feb 2020, 22:27.
Last edited by Bunuel on 11 Feb 2020, 23:10, edited 1 time in total.
Renamed the topic and edited the question.
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What is the greatest positive integer x such that 24^x is a factor of  [#permalink]

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New post 11 Feb 2020, 23:46
We basically need to determine the highest power of 24 = 2³ * 3 in 100!

Let us determine the highest power of 2 in 100! :
[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64] = 97

where [n] refers to the greatest integer less than or equal to x

Thus, highest power of 2³ in 100! = [97/3] = 32

Basically we are grouping 3 2s at a time from the 97 2s available with us, and hence we have 32 such groups

Let us determine the highest power of 3 in 100! :
[100/3] + [100/9] + [100/27] + [100/81] = 48

Thus, there are 32 groups of 3 2s and 48 3s

We need one group of 3 2s and one 3 to form a 24. Hence, we can do that only 32 times.

Hence, highest power of 24 in 100! is 32

Answer A

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Re: What is the greatest positive integer x such that 24^x is a factor of  [#permalink]

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New post 11 Feb 2020, 23:46
kapil1 wrote:
What is the greatest positive integer x such that 24^x is a factor of 100!?​

A. 32​
B. 33​
C. 46​
D. 47​
E. 48​



So we look at the times 24 is there in 100!. For this we will check the number of times the prime numbers are there..

Now \(24=2^3*3\)

Number of \(2^3\)
\(\frac{100}{2}+\frac{100}{2^2}+\frac{100}{2^3}+\frac{100}{2^4}+\frac{100}{2^5}+\frac{100}{2^6}=50+25+12+6+3+1=97\)
We take only the integer part
Since we are looking for number of \(2^3\), it will be equal to \(\frac{97}{3}=32\)

Number of \(3\)
\(\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}=33+11+3+1=48\)
We take only the integer part

So there are THIRTY TWO 2^3 and FORTY EIGHT 3.
We will have the lesser of two, that is 32, number of 24, so x=32

A
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Re: What is the greatest positive integer x such that 24^x is a factor of  [#permalink]

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New post 16 Feb 2020, 05:44
kapil1 wrote:
What is the greatest positive integer x such that 24^x is a factor of 100!?​

A. 32​
B. 33​
C. 46​
D. 47​
E. 48​


Since 24 = 2^3 x 3^1, there will be fewer 2^3’s than 3^1’s in 100!, and therefore, we need to determine the number of 2^3’s in 100!. However, we need to first determine the number of 2s in 100!.

100/2 = 50

100/2^2 = 25

100/2^3 = 12

100/2^4 = 6

100/2^5 = 3

100/2^6 = 1

100/2^7 = 0

Thus, the number of twos is 50 + 25 + 12 + 6 + 3 + 1 = 97.

We want to determine the number of 2^3’s there are in 100!, so we divide by 3. Since 97/3 = 32.2, the number of 2^3’s in 100! is 32. That is, the largest value of x such that 24^x is a factor of 100! is 32.

Answer: A
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Re: What is the greatest positive integer x such that 24^x is a factor of  [#permalink]

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New post 16 Feb 2020, 11:51
kapil1 wrote:
What is the greatest positive integer x such that 24^x is a factor of 100!?​

A. 32​
B. 33​
C. 46​
D. 47​
E. 48​


\(24^x = (2^3*3)^x = 2^{3x}* 3^x\).

We want to count how many multiples of 24 are in 100!. This will be either limited by the number of multiples of two's in 100! or the number of multiples of three's. Let's start from the multiples of two's, there is a nice way to count the multiples of two's using layers.

1st layer: Multiples of twos from 2 to 100. 100/2 = 50 numbers. (The didactic formula is (High - Low)/2 + 1 = (100 - 2)/2 + 1 but we can do 100/2 since we're starting from the first multiple of two).
2nd layer: Multiples of fours from 4 to 100. 100/4 = 25 numbers. Note this only adds 25 to the count of multiples of twos since we already included their first multiple of two in the first layer.
3rd layer: Multiples of eights from 8 to 100. We actually count 8 to 96, which is 96/8 = 12 numbers.
4th layer: Multiples of sixteen. 16 to 96. 6 numbers.
5th layer: 32, 64, 96. 3 numbers
6th layer: 64. 1 number.

50 + 25 + 12 + 6 + 3 + 1 = 75 + 12 + 10 = 97 multiples of two. However since 97/3 = (96 + 1)/3 = 32 with a remainder of 1, we can only insert 32 multiples of twenty-four with this many multiples of twos, since each twenty-four needs 3 multiples of twos.

Let us repeat again to find the number of multiples of three's:
1st layer: 3 to 99. 33 numbers.

We can stop here. We know there are more than 32 multiples of three's and the limiting condition here is the number of multiples of two's. Thus we can only find 32 multiples of twenty-fours.

Ans: A
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Re: What is the greatest positive integer x such that 24^x is a factor of   [#permalink] 16 Feb 2020, 11:51
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