MGMAT Number Theory

[kool_sunny]

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 will follow soon

[e.cartman]

n=4x+1=5y+3
Remainder 1 when div by 4 and 3 when div by 5.
13+33=46. B

[kool_sunny]

Originally Posted by e.cartman

n=4x+1=5y+3
Remainder 1 when div by 4 and 3 when div by 5.
13+33=46. B

You really rock e.cartman....

OA is B

But can please you explain in more detail your strategy?

[e.cartman]

I just listed multiples of 4 and 5 and used brute force. But I've heard of people use chinese remainder theorem for faster results. But I guess it needs practice. If somebody knows simple steps please share.

[ranjeet_1975]

Originally Posted by e.cartman

n=4x+1=5y+3
Remainder 1 when div by 4 and 3 when div by 5.
13+33=46. B

why you have chosen 13 and 33 ? It may be 5 and 8, or 9 and 13 etc.
please explain.

[vinhan]

let n = 4k +1
this n when divided into group of 5, then rem is 3.
i.e. 4k-2 is exactly divisible by 5.

the least values of k are 3 & 6.

the sum is 13 + 33 = 46

[Retake_GMAT]

5K +3 =4m +1
5k-4m=-2
5k = -2 + 4m
K = (-2 +4m) /5
If m=3, k=10/5 =2
If m=8, k=30/5 =6
Hence numbers are : 13 and 33, hence sum =13+ 33 =46

[12rk34]

n= 4m1 + 1 & n = 5m2 + 3 (m1 & m2 are number of groups respectively.)
Both are A.P. with common difference of 4 and 5 respectively.
So putting m1 = 1, 2, 3, 4 ..
we get n = 5, 9, 13, 17, 21, 25, 29, 33, 37, ...
Putting m2 = 1, 2, 3, ...
n = 8, 13, 18, 23, 28, 33, 38,...
Smallest possible values common to both A. P. are 13 & 33.
Sum is 46. Hence B.

[ankit0uc]

I think question is wrongly worded -going by the above wording answer should be 13+13=26

Instead it should be What is the possible sum of the two smallest possible values of n?