If K is a positive integer less than 10 and N = 4,321 + K,

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If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

(1) N is divisible by 3.
(2) N is divisible by 7.



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by SampathKp » Sun Dec 29, 2019 8:07 pm
BTGmoderatorDC wrote:If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

(1) N is divisible by 3.
(2) N is divisible by 7.



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Given that N = 4321 + K and and K is less than 10.

From (1) N is divisible by 3 but as K is less than 10 . so it can have upto 3 values for which N can be divisible by 3. Clearly this Information is NOT sufficient.

From (2) N is divisible by 7. Dividing 4321 by 7 we get a remainder of 2. So if we add another 5 to 4321 then it will be divisible by 7. K =5. Next +ve integer after 5 that has to be added to 4321 for it to be divisible by 7 is 12. But given K is less than 10. hence K=5 and This Information is Sufficient.

Answer is B

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by Brent@GMATPrepNow » Mon Dec 30, 2019 6:48 am
BTGmoderatorDC wrote:If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

(1) N is divisible by 3.
(2) N is divisible by 7.
Target question: What is the value of K?

Given: K is a positive integer less than 10 and N = 4,321 + K
If K is a positive integer less than 10, then K = 1, 2, 3, 4, ... or 9
So, 4321 + K (aka N) can have 9 possible values.
In other word, N = 4322, 4323, 4324, ..., 4330 (9 consecutive integers)

Statement 1: N is divisible by 3
IMPORTANT RULE: Among a set of consecutive integers, every 3rd integer will be divisible by 3.
Likewise, every 4th integer will be divisible by 4.
Every 5th integer will be divisible by 5.
Etc.
So, in a set of consecutive integers, about 1/3 of them will be divisible by 3.

Since N can be any of the 9 consecutive integers from 4322 to 4330, and since 1/3 of these 9 integers are divisible by 3, we can conclude that N could have 3 possible values.
Aside: We need not determine what these 3 possible values are, but if we were to find them, we'd see that N could equal 4323, 4326 or 4329
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: N is divisible by 7
Using our rule from earlier, about 1/7 of the 9 integers are divisible by 7, so in this case, it's possible that there's 1 value for N or possibly 2 values.
So, we need to check.
We'll look for a multiple of 7 among the possible values of N (4322, 4323, 4324, ..., 4330)
We know that 4200 is divisible by 7
So, 4270 is divisible by 7 (added 70 to 4200)
Which means 4340 is divisible by 7 (added 70 to 4270)
Oh, we've gone to far.
4333 is divisible by 7 (subtracted 7 from 4340)
4326 is divisible by 7 (subtracted 7 from 4333) BINGO

Since every 7th integer is divisible by 7, we can see that N could equal ... 4212, 4219, 4326, 4333, 4340,...
Since only one of these values is in the range of possible values of N (4322, 4323, 4324, ..., 4330), we can be certain that N = 4326
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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