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N = 10^40 + 2^40. 2^k is a divisor of N,

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

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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 \(k-2\) ?

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,  [#permalink]

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New post Updated on: 01 Apr 2017, 22:28
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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 \(k-2\) ?

A) 38
B) 39
C) 40
D) 41
E) 42

*kudos for all correct solutions


Hi

Good question.

\(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\)

\(5^{any-power}\) 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, 10:35.
Last edited by vitaliyGMAT on 01 Apr 2017, 22:28, edited 1 time in total.
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N = 10^40 + 2^40. 2^k is a divisor of N,  [#permalink]

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New post Updated on: 01 Apr 2017, 22:54
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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 \(k-2\) ?

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 k-1 = 39

Hence option B is correct
Hit Kudos if you liked it 8-)

Originally posted by 0akshay0 on 01 Apr 2017, 10:47.
Last edited by 0akshay0 on 01 Apr 2017, 22:54, edited 1 time in total.
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Re: N = 10^40 + 2^40. 2^k is a divisor of N,  [#permalink]

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New post 01 Apr 2017, 13:05
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vitaliyGMAT wrote:
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.


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

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New post 02 Apr 2017, 02:46
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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.

k-2 = 41-2 =39

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

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New post 02 Apr 2017, 07:17
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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 \(k-2\) ?

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 = 25
5^3 = 125
5^4 = 625
5^5 = 3125
5^6 = XXX25etc....

So......
N = 10^40 + 2^40
= 2^40(5^40 + 1)
= 2^40(XXXX25 + 1) [aside: XXXX25 denotes some number ending in 25]
= 2^40(XXXX26)
= 2^40[2(XXXX3)] [Since XXX26 is EVEN, we can factor out a 2]
= 2^41[XXXX3]

Since XXXX3 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 k-2?
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,  [#permalink]

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Re: N = 10^40 + 2^40. 2^k is a divisor of N, &nbs [#permalink] 23 Jul 2018, 03:09
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