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Manhattan Question Set # 13

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Manhattan Question Set # 13

by richachampion » Wed Oct 12, 2016 7:38 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
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by [email protected] » Wed Oct 12, 2016 9:43 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 richachampion » Wed Oct 12, 2016 9:47 am
[email protected] wrote: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

GMAT assassins aren't born, they're made,
Rich
Thanks sir. I used the same method.
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:30 am
Let's say a and b are integers whose values we don't yet know. We know that

4*a + 1 = n

5*b + 3 = n

So we must have

4a + 1 = 5b + 3

4a = 5b + 2

Since we want the two smallest positive values of n, we want the two smallest positive values of b, i.e. the first two multiples of 5 that are 2 less than a multiple of 4. These can only be even multiples of 5: 10 works, 20 doesn't, 30 does, so our two solutions are

5b = 10, b = 2, n = 13
and
5b = 30, b = 6, n = 33

and the solution is 13 + 33 = 46.
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:33 am
A neat way of solving this would be cheating the answer choices a bit. Since both solutions are of the form

5*(some integer) + 3, we know that

(smallest n) + (next smallest n) must be

5*x + 3 + 5*y + 3, with x and y nonnegative integers. So the solution must be

5*(x + y) + 6

Only 46 and 86 are possible. Trying 46 as the sum of two numbers that end in 3, we get 13 + 33, which works, so this is our choice!
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