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Manhattan Question Set # 18

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by richachampion » Wed Oct 12, 2016 7:42 am
A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33
B. 46
C. 49
D. 53
E. 86

OA: B
R I C H A,
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by [email protected] » Wed Oct 12, 2016 10:12 am
Hi richachampion,

You would likely find it easiest to 'brute force' this question (simply write down enough of the possibilities until you either spot the pattern involved or have the exact answer on your pad).

Equal groups of 4 with 1 left over COULD be... 5, 9, 13, 17, 21, 25, 29, 33.....
Equal groups of 5 with 3 left over COULD be... 8, 13, 18, 23, 28, 33....

The two SMALLEST values that fit BOTH groups are 13 and 33. We're asked for the sum of those values...

Final Answer: B

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by Brent@GMATPrepNow » Thu Oct 13, 2016 7:14 am
richachampion wrote:A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33
B. 46
C. 49
D. 53
E. 86
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

-----------------------------

n students can be divided into equal groups of 4 with 1 student left over
So, n divided by 4 leaves a remainder of 1.
So, the possible values of n are: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, etc.

Aside: It's unclear whether n could equal 1 here. Can we say that 1 student can be divided into equal groups of 4 with 1 student left over? Even 5 students might night be okay, since we can't really say that 5 students can be divided into equal groups of 4 with 1 student left over. Anyhoo, it doesn't really matter since those two values (1 and 5) don't come into play.

n students can be divided into equal groups of 5 with 3 students left over
So, n divided by 5 leaves a remainder of 3.
So, the possible values of n are: 3, 8, 13, 18, 23, 28, 33, 38, etc.

So, the two smallest values of n that satisfy BOTH conditions are 13 and 33.

13 + 33 = 46
Answer: B

RELATED VIDEO
- Introduction to Remainders: https://www.gmatprepnow.com/module/gmat ... /video/842
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Thu Oct 13, 2016 7:14 am
richachampion wrote:A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33
B. 46
C. 49
D. 53
E. 86
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

-----------------------------

n students can be divided into equal groups of 4 with 1 student left over
So, n divided by 4 leaves a remainder of 1.
So, the possible values of n are: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, etc.

Aside: It's unclear whether n could equal 1 here. Can we say that 1 student can be divided into equal groups of 4 with 1 student left over? Even 5 students might not be okay, since we can't really say that 5 students can be divided into equal groups of 4 with 1 student left over. Anyhoo, it doesn't really matter since those two values (1 and 5) don't come into play.

n students can be divided into equal groups of 5 with 3 students left over
So, n divided by 5 leaves a remainder of 3.
So, the possible values of n are: 3, 8, 13, 18, 23, 28, 33, 38, etc.

So, the two smallest values of n that satisfy BOTH conditions are 13 and 33.

13 + 33 = 46
Answer: B

RELATED VIDEO
- Introduction to Remainders: https://www.gmatprepnow.com/module/gmat ... /video/842
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion