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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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A. 3
B. 5
C. 6
D. 7
E. 8
_________________
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Bunuel
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
Option E)
\(\frac{11}{(2^1)} = 5\)
\(\frac{11}{(2^2)} = 2\)
\(\frac{11}{(2^3)} = 1\)
Y \(= 5 + 2 + 1 = 8\)
\(\frac{11}{(2^1)} = 5\)
\(\frac{11}{(2^2)} = 2\)
\(\frac{11}{(2^3)} = 1\)
Y \(= 5 + 2 + 1 = 8\)
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
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
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
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
JeffTargetTestPrep
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Joined: 04 Mar 2011
Last visit: 05 Jan 2024
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Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
Bunuel
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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Bunuel
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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JeffTargetTestPrep
Interesting, Could you please explain the understanding behind this method. I'm interested. Thanks
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