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N = 10^40 + 2^40. 2^k is a divisor of N,
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01 Apr 2017, 07:07
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\(N = 10^{40} + 2^{40}\). \(2^k\) 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 \(k2\) ? A) 38 B) 39 C) 40 D) 41 E) 42 *kudos for all correct solutions
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N = 10^40 + 2^40. 2^k is a divisor of N,
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Updated on: 01 Apr 2017, 21:54
GMATPrepNow wrote: \(N = 10^{40} + 2^{40}\). \(2^k\) 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 \(k2\) ?
A) 38 B) 39 C) 40 D) 41 E) 42
*kudos for all correct solutions \(N = 10^{40} + 2^{40}\) \(N = (5*2)^{40} + 2^{40}\) \(N = 2^{40}(5^{40}+1)\) 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 k1 = 39 Hence option B is correct Hit Kudos if you liked it
Originally posted by 0akshay0 on 01 Apr 2017, 09:47.
Last edited by 0akshay0 on 01 Apr 2017, 21:54, edited 1 time in total.




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N = 10^40 + 2^40. 2^k is a divisor of N,
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Updated on: 01 Apr 2017, 21:28
GMATPrepNow wrote: \(N = 10^{40} + 2^{40}\). \(2^k\) 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 \(k2\) ?
A) 38 B) 39 C) 40 D) 41 E) 42
*kudos for all correct solutions Hi Good question. \(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\) \(5^{anypower}\) will always be odd, hence \(5^{40} + 1\) will be even. The question is: can we get multiple of 4? There are two possible scenarious for a number ending in 5. Last two digits can be either 75 or 25. In the first case if we add 1 number will be divisible by 4, in second no. 5 to the power greater than 1 will always have last two digits 25. 25 + 1 = 26 not divisible by 4. Answer to the question is no. Hence \(max\) power of \(2\) in \(N\) is \(41\). \(k=41\), > \(k  2 = 39\). Answer B.
Originally posted by vitaliyGMAT on 01 Apr 2017, 09:35.
Last edited by vitaliyGMAT on 01 Apr 2017, 21:28, edited 1 time in total.



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Re: N = 10^40 + 2^40. 2^k is a divisor of N,
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01 Apr 2017, 12:05
vitaliyGMAT wrote: Good question.
\(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\)
\(5^{anypower}\) 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 = 255^3 = 1 255^4 = 6 255^5 = 31 25 etc.... So, 5^40 + 1 = ????? 25 + 1 = ?????26 And when we divide ?????26 by 2, the quotient is ?????? 3Cheers, Brent
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Re: N = 10^40 + 2^40. 2^k is a divisor of N,
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02 Apr 2017, 01:46
Really Tricky question, Brent!!
N = 10^40 + 2^40 = (2^40)(5^40) + (2^40)= (2^40) (5^40 + 1)
2^k is divisor of N
(2^40) (5^40 + 1)/(2^k).....it means that 2^40........k= 40. However, when apply k=40 in 2^(k+1), it is still advisable because the term (5^40 + 1) hides another 2 as (5^40 + 1)= (number with unit digit 5+ 1) = number with unit digit of 6.
Therefore k=41 So make all conditions above works.
k2 = 412 =39
Answer:B



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Re: N = 10^40 + 2^40. 2^k is a divisor of N,
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02 Apr 2017, 06:17
GMATPrepNow wrote: \(N = 10^{40} + 2^{40}\). \(2^k\) 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 \(k2\) ?
A) 38 B) 39 C) 40 D) 41 E) 42
IMPORTANT: we need to recognize that 5^b will end in 25 for all integer values of b greater than 1. For example, 5^2 = 255^3 = 1 255^4 = 6 255^5 = 31 25 5^6 = XXX 25etc.... So...... N = 10^40 + 2^40 = 2^40(5^40 + 1) = 2^40(XXXX 25 + 1) [aside: XXXX25 denotes some number ending in 25]= 2^40(XXXX 26) = 2^40[2(XXXX 3)] [Since XXX26 is EVEN, we can factor out a 2]= 2^41[XXXX 3] Since XXXX 3 is an ODD number, we cannot factor any more 2's out of it. This means that 2^41 IS a factor of N, but 2^42 is NOT a factor of N. In other words, k = 41 What is the value of k2? Since k = 41, we can conclude that k  2 = 41  2 = 39 Answer: Cheers, Brent
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Re: N = 10^40 + 2^40. 2^k is a divisor of N,
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29 Oct 2019, 07:52
0akshay0 wrote: GMATPrepNow wrote: \(N = 10^{40} + 2^{40}\). \(2^k\) 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 \(k2\) ?
A) 38 B) 39 C) 40 D) 41 E) 42
*kudos for all correct solutions \(N = 10^{40} + 2^{40}\) \(N = (5*2)^{40} + 2^{40}\) \(N = 2^{40}(5^{40}+1)\) 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 k1 = 39 Hence option B is correct Hit Kudos if you liked it iN YOUR SOLUTION YOU MEAN K2 = 39 RIGHT ? AND NOT k1




Re: N = 10^40 + 2^40. 2^k is a divisor of N,
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29 Oct 2019, 07:52






