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# If n is the product of all the integers from 5 to 20, inclusive, what

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If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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11 Nov 2016, 01:26
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If n is the product of all the integers from 5 to 20, inclusive, what is the greatest integer k for which 2^k is a factor of n ?

(A) 11

(B) 12

(C) 13

(D) 15

(E) 16

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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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11 Nov 2016, 02:25
The question is simply asking " how many 2's are there from 5 to 20.
There are totally 15 2's.
Ans D
I had to calculate manually (90 seconds to complete ).. any easier method is welcome.
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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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14 Nov 2016, 01:28
The questions asks you about the number of 2, from n=(5)(6)….(19)(20)=2k(int). Then, you only have to find even numbers from (5)(6)..(19)(20), and only 6,8,10,12,14,16,18,20 are even numbers. The numbers of 2 for each number are 6=(2)(3), 8=23, 10=(2)(5), 12=(22)(3), 14=(2)(7), 16=24, 18=(2)(32), 20=(22)(5), and 2 becomes 2, 23, 2, 22, 2, 24, 2, 22. Hence the number of 2 is 1+3+1+2+1+4+2+1=15.

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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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14 Nov 2016, 02:21
20/2=10
10/2=5
Ans:10+5=15
ans:D

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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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14 Nov 2016, 08:16
2
MathRevolution wrote:
If n is the product of all the integers from 5 to 20, inclusive, what is the greatest integer k for which 2^k is a factor of n ?

(A) 11

(B) 12

(C) 13

(D) 15

(E) 16

We are given that n is the product of all the integers from 5 to 20, inclusive, and we must determine the greatest integer k for which 2^k is a factor of n. Although there is no direct shortcut here to determine the number of factors of 2 within the product of the integers from 5 to 20 inclusive, there is a shortcut to determine the number of factors of 2 in the product of integers from 1 to 20 inclusive (or 20!). So to determine the number of 2s in the product of the integers from 5 to 20, we can determine the number of 2s in the product of the integers from 1 to 20 inclusive and subtract the number of 2s in the product of the numbers from 1 to 4 inclusive.

To determine the number of 2s within 20!, we can use the following shortcut in which we divide 20 by 2, then divide the quotient of 20/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

20/2 = 10 (we can ignore the remainder)

10/2 = 5

5/2 = 2 (we can ignore the remainder)

2/2 = 1

Since 1/2 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 2 within 20!.

Thus, there are 10 + 5 + 2 + 1 = 18 factors of 2 within 20!

Now we need to subtract the number of 2s in the product of the numbers from 1 to 4 inclusive. There are three factors of 2 in that product; thus, there are 18 - 3 = 15 factors of 2 within the product of the integers from 5 to 20 inclusive.

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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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28 Jan 2018, 05:37
15 2's
this is small set , we can count
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Re: If n is the product of all the integers from 5 to 20, inclusive, what  [#permalink]

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10 Mar 2019, 16:47
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Re: If n is the product of all the integers from 5 to 20, inclusive, what   [#permalink] 10 Mar 2019, 16:47
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# If n is the product of all the integers from 5 to 20, inclusive, what

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