November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. November 17, 2018 November 17, 2018 09:00 AM PST 11:00 AM PST Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.
Author 
Message 
TAGS:

Hide Tags

CEO
Joined: 11 Sep 2015
Posts: 3120
Location: Canada

N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
01 Apr 2017, 07:07
Question Stats:
31% (01:54) correct 69% (01:52) wrong based on 131 sessions
HideShow timer Statistics
\(N = 10^{40} + 2^{40}\). \(2^k\) 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Brent Hanneson – GMATPrepNow.com



Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98

N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
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.



Senior Manager
Joined: 19 Apr 2016
Posts: 274
Location: India
GMAT 1: 570 Q48 V22 GMAT 2: 640 Q49 V28
GPA: 3.5
WE: Web Development (Computer Software)

N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
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.



CEO
Joined: 11 Sep 2015
Posts: 3120
Location: Canada

Re: N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
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
_________________
Brent Hanneson – GMATPrepNow.com



SVP
Joined: 26 Mar 2013
Posts: 1879

Re: N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
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



CEO
Joined: 11 Sep 2015
Posts: 3120
Location: Canada

Re: N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
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
_________________
Brent Hanneson – GMATPrepNow.com



NonHuman User
Joined: 09 Sep 2013
Posts: 8772

Re: N = 10^40 + 2^40. 2^k is a divisor of N,
[#permalink]
Show Tags
23 Jul 2018, 02:09
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: N = 10^40 + 2^40. 2^k is a divisor of N, &nbs
[#permalink]
23 Jul 2018, 02:09






