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

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

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08 Dec 2016, 11:58
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15% (low)

Question Stats:

75% (00:49) correct 25% (01:01) wrong based on 162 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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Re: What is the greatest value of y such that 2^y is a factor of 11! ?  [#permalink]

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08 Dec 2016, 19:07
1
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]

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08 Dec 2016, 23:12
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]

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08 Dec 2016, 23:32
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]

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08 Dec 2016, 23:34
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]

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12 Dec 2016, 17:14
1
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!.

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

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21 Mar 2018, 04:08
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]

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20 Jul 2018, 08:27
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!.

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! ? &nbs [#permalink] 20 Jul 2018, 08:27
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