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k = 2^n + 7, where n is an integer greater than 1. If k is divisible by 9,

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by BTGModeratorVI » Tue Mar 31, 2020 5:07 am

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k = 2^n + 7, where n is an integer greater than 1. If k is divisible by 9, which of the following MUST be divisible by 9?

A. 2^n - 8
B. 2^n - 2
C. 2^n
D. 2^n + 4
E. 2^n + 5

Answer: B
Source: Economist GMAT
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Source: — Problem Solving |

Hi All,

We're told that K = (2^N) + 7, where N is an integer greater than 1 AND K is divisible by 9. We're asked which of the following MUST be divisible by 9? This is a great 'concept question', meaning that you don't actually have to do much math to answer this question if you recognize the concept(s) involved.

If a particular number is divisible by another number, then you can 'count' up to the next number that is also divisible. For example...

21 is divisible by 3.... meaning that the NEXT number that is divisible by 3 will be "3 away"....
re: 21 + 3 = 24

In that same way, we can word backwards to find the PRIOR number that is also divisible by 3. It's "3 before"....
re: 21 - 3 = 18

In this question, we know that (2^N) + 7 is divisible by 9... so the next number and the prior number that are also divisible by 9 are....
(2^N) + 7 + 9 = (2^N) + 16
(2^N) + 7 - 9 = (2^N) -2

Final Answer: B

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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BTGModeratorVI wrote:
Tue Mar 31, 2020 5:07 am
k = 2^n + 7, where n is an integer greater than 1. If k is divisible by 9, which of the following MUST be divisible by 9?

A. 2^n - 8
B. 2^n - 2
C. 2^n
D. 2^n + 4
E. 2^n + 5

Answer: B
Source: Economist GMAT
Since k/9 = integer, (2^n + 7)/9 = integer. Since 2^n + 7 is a multiple of 9, we can add or subtract any multiple of 9 to 2^n + 7 and that expression will still be divisible by 9. Thus:

2^n + 7 - 9 = 2^n - 2 MUST be divisible by 9.

Answer: B

Scott Woodbury-Stewart
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