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tricky exponent question - N = 10^40 + 2^40

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by Brent@GMATPrepNow » Sat Apr 01, 2017 7:08 am
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

Answer: B

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by Mo2men » Sat Apr 01, 2017 2:37 pm
Brent@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

Answer: B

Source: www.gmatprepnow.com
Difficulty level: 700+
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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by Brent@GMATPrepNow » Sun Apr 02, 2017 6:16 am
Brent@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: B

Cheers,
Brent
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