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

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Joined: 11 Sep 2015
Posts: 4340
N = 10^40 + 2^40. 2^k is a divisor of N,  [#permalink]

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01 Apr 2017, 07:07
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95% (hard)

Question Stats:

32% (01:50) correct 68% (02:01) wrong based on 177 sessions

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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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Updated on: 01 Apr 2017, 21:54
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1
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

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

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Updated on: 01 Apr 2017, 21:28
1
2
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 $$41$$.

$$k=41$$, ----> $$k - 2 = 39$$.

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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Posts: 4340
Re: N = 10^40 + 2^40. 2^k is a divisor of N,  [#permalink]

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01 Apr 2017, 12:05
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Top Contributor
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 $$41$$.

$$k=41$$, ----> $$k - 2 = 39$$.

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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02 Apr 2017, 01:46
1
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

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

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02 Apr 2017, 06:17
1
1
Top Contributor
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

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

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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 $$k-2$$ ?

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

iN YOUR SOLUTION YOU MEAN K-2 = 39 RIGHT ? AND NOT k-1
Re: N = 10^40 + 2^40. 2^k is a divisor of N,   [#permalink] 29 Oct 2019, 07:52
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