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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