When a natural number N is successively divided by 5,6,7

[Gmat_mission]

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

[spoiler]OA=D[/spoiler]

This is a really interesting question. First, I thought that the remainders were the same, but they were not. I couldn't solve this PS question. May anyone gives me some help? Please, I'd be thankful.

[Jeff@TargetTestPrep]

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

[swerve]

5 6 7 8
1 2 4 5

Leave the top right- most number 8.
Start with bottom right-most number 5
5*7+4=39
39*6+2=236
236*5+1=1181

This is the number required.
Now, do the successive division in the reverse order.
The sum of the remainders is 11

Hence, the correct option is D.