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What is the greatest value of y such that 2^y is a factor of 11! ?

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Bunuel
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What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   08 Dec 2016, 12:58
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E

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73% (00:47) correct 27% (00:56) wrong based on 179 sessions
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What is the greatest value of y such that 2^y is a factor of 11! ?

A. 3
B. 5
C. 6
D. 7
E. 8
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E
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rohit8865
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   08 Dec 2016, 20:07
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Bunuel wrote:
What is the greatest value of y such that 2^y is a factor of 11! ?

A. 3
B. 5
C. 6
D. 7
E. 8



11/2=5
11/2^2=2
11/2^3=1

we will not calculate 11/2^4 as 2^4>11

total= 5+2+1=8

Ans E
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   09 Dec 2016, 00:12
Option E)

\(\frac{11}{(2^1)} = 5\)
\(\frac{11}{(2^2)} = 2\)
\(\frac{11}{(2^3)} = 1\)

Y \(= 5 + 2 + 1 = 8\)
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   09 Dec 2016, 00:32
Greatest value of y such that 2^y is a factor of 11!

11! = 1*2*3*4*5*6*7*8*9*10*11 which is same also 11! = 1*2*3*2*2*5*2*3*7*2*2*2*9*2*5*11

Count all occurrence of 2 in the expression = 8

The largest power of 2 that will still result in a factor of 11! is 2^8
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   09 Dec 2016, 00:34
Greatest value of y such that 2^y is a factor of 11!

11! = 1*2*3*4*5*6*7*8*9*10*11 which is same also 11! = 1*2*3*2*2*5*2*3*7*2*2*2*9*2*5*11

Count all occurrence of 2 in the expression = 8

The largest power of 2 that will still result in a factor of 11! is 2^8

Ans E
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   12 Dec 2016, 18:14
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Bunuel wrote:
What is the greatest value of y such that 2^y is a factor of 11! ?

A. 3
B. 5
C. 6
D. 7
E. 8


To determine the number of factors of 2 within 11!, we can use the following shortcut in which we divide 11 by 2, and then divide the quotient of 11/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

11/2 = 5 (we can ignore the remainder)

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 11!.

Thus, there are 5 + 2 + 1 = 8 factors of 2 within 11!.

Answer: E
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What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   21 Mar 2018, 05:08
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Bunuel wrote:
What is the greatest value of y such that 2^y is a factor of 11! ?

A. 3
B. 5
C. 6
D. 7
E. 8


11! means \(1*2*3*4*5*6*7*8*9*10*11\)

Find all the factors of \("2"\)

\((2^1)*(2^2)*(2^1*3)*(2^3)*(2^1*5)\)

\(2^{1+2+1+3+1}\)

\(2^8\)

Hence (E)
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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   20 Jul 2018, 09:27
JeffTargetTestPrep wrote:
Bunuel wrote:
What is the greatest value of y such that 2^y is a factor of 11! ?

A. 3
B. 5
C. 6
D. 7
E. 8


To determine the number of factors of 2 within 11!, we can use the following shortcut in which we divide 11 by 2, and then divide the quotient of 11/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

11/2 = 5 (we can ignore the remainder)

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 11!.

Thus, there are 5 + 2 + 1 = 8 factors of 2 within 11!.

Answer: E

Interesting, Could you please explain the understanding behind this method. I'm interested. Thanks

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Re: What is the greatest value of y such that 2^y is a factor of 11! ? [#permalink] New post   29 Jul 2022, 06:22
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