enigma123 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
OA is A and this is how I arrived at.
Let say n = 4x+1 and n = 5y+3 -----> From the question Stem
n=4x+1 n = 5y+3
5 8
9 13
13 18
17 23
25 33
29 38
33 43
I get these above values by putting the same values for x and y. Is my concept correct?
Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.
A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9,
13, 17, 21, 25, 29,
33, 37, ... (basically an evenly spaced set with common difference of 4)
A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8,
13, 18, 23, 28,
33, 38, ... (basically an evenly spaced set with common difference of 5)
Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.
Answer: B.
Else you can derive general formula based on n=4q+1 and n=5p+3.
Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.
Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.
Answer: B.
For more about this concept see:
https://gmatclub.com/forum/manhattan-re ... ml#p721341https://gmatclub.com/forum/when-positiv ... ml#p722552https://gmatclub.com/forum/when-the-pos ... l#p1028654Hope it helps.
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