Gmat_mission wrote:When a natural number N is successively divided by 5,6,7,8 the remainders are 1,2,4,5. What will be the sum of the remainders if the order of the division is reversed?

A) 16
B) 14
C) 12
D) 11
E) 9

Since the remainder is 1 when N is divided by 5, N = 5Q + 1 for some integer Q. Now, if we take Q and divide it by 6, we will have a remainder 2. So Q = 6P + 2 for some integer P. Now if we take P and divide it by 7, we will have a remainder 4. So P = 7S + 4 for some integer S. Finally, if we take S and divide it by 8, we will have a remainder 5. So S = 8T + 5 for some integer T. In summary,

N = 5Q + 1 where

Q = 6P + 2 where

P = 7S + 4 where

S = 8T + 5 for some integer T

We can reveal what N could be by letting T to be any non-negative integer. Of course, the smallest non-negative integer is 0, so let's let T = 0, then we will have:

S = 8(0) + 5 = 5 and

P = 7(5) + 4 = 39 and

Q = 6(39) + 2 = 236 and

N = 5(236) + 1 = 1181

So the smallest value N could be is 1181 and divide it successively by 8, 7, 6 and 5:

1181 / 8 = 147 R 5

147 / 7 = 21 R 0

21 / 6 = 3 R 3

3 / 5 = 0 R 3

So the sum of the new remainders is 5 + 0 + 3 + 3 = 11.

Answer: D