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01 Apr 2017, 08:07
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B
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Difficulty:
95%
(hard)
Question Stats:
42% (01:53) correct
58%
(01:59)
wrong
based on 367
sessions
History
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Not Attempted Yet
\(N = 10^{40} + 2^{40}\). \(2^k\) is a divisor of \(N\), but \(2^{k+1}\) is not a divisor of \(N\). If k is a positive integer, what is the value of \(k-2\) ?
A) 38
B) 39
C) 40
D) 41
E) 42
A) 38
B) 39
C) 40
D) 41
E) 42
0akshay0
Joined: 19 Apr 2016
Last visit: 14 Jul 2019
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Location: India
Schools: Jindal '20 Kelley '20 Broad '20 Carey '20 Mays '20
GMAT 1: 570 Q48 V22
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WE:Web Development (Computer Software)
GMATPrepNow
\(N = 10^{40} + 2^{40}\). \(2^k\) is a divisor of \(N\), but \(2^{k+1}\) is not a divisor of \(N\). If k is a positive integer, what is the value of \(k-2\) ?
A) 38
B) 39
C) 40
D) 41
E) 42
*kudos for all correct solutions
\(N = 10^{40} + 2^{40}\)A) 38
B) 39
C) 40
D) 41
E) 42
*kudos for all correct solutions
\(N = (5*2)^{40} + 2^{40}\)
\(N = 2^{40}(5^{40}+1)\) -----------------I
now \(5^{anything}\) results in a number that ends with 5 as the unit digit so the last two digits of
\((5^{40}+1)\) will always be 26 which is divisible by 2 only once
so equation I becomes
\(N = 2^{40}*2*something\)
\(N = 2^{41}*something\)
Therefore k = 41 and k-2 = 39
Hence option B is correct
01 Apr 2017, 13:05
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vitaliyGMAT
Good question.
\(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\)
\(5^{any-power}\) will always be odd, hence \(5^{40} + 1\) will be even.
The question is: can we get multiple of 4? Last digit of \(5^{40} + 1\) will be 6. When we divide that huge number \(...........6\) by \(2\) we'll get last digit \(3\) (odd) and the answer to the question is no.
Hence \(max\) power of \(2\) in \(N\) is \(41\).
\(k=41\), ----> \(k - 2 = 39\).
Answer B.
\(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\)
\(5^{any-power}\) will always be odd, hence \(5^{40} + 1\) will be even.
The question is: can we get multiple of 4? Last digit of \(5^{40} + 1\) will be 6. When we divide that huge number \(...........6\) by \(2\) we'll get last digit \(3\) (odd) and the answer to the question is no.
Hence \(max\) power of \(2\) in \(N\) is \(41\).
\(k=41\), ----> \(k - 2 = 39\).
Answer B.
That's a solid solution. However, I'll point out one incorrect statement, in case it helps out with a similar question in the future.
You're correct to say that the last digit of 5^40 + 1 will be 6.
However, knowing that the units digit of a number is 6, does not necessarily mean that, when we divide that number by 2, the units digit of the quotient will be 3.
For example, 1056 divided by 2 equals 528
Likewise, 96 divided by 2 equals 48
So, to remove any uncertainty, we must recognize that 5^b will end in 25 for all integer values of b greater than 1.
For example, 5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125 etc....
So, 5^40 + 1 = ?????25 + 1 = ?????26
And when we divide ?????26 by 2, the quotient is ??????3
Cheers,
Brent
