Subanta
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
Good question.
By successive it means that we divide N first by 5 and obtain a quotient 'a' and remainder of 1. Now we divide 'a' by 6 and obtain a quotient of b and remainder of 2. Similarly, dividing 'b' by 7 gives us a quotient 'c' and remainder of 4. Finally, dividing 'c' by 8 gives us a quotient of 'd' and remainder of 5.
For any number N when divided by another number S, we get a quotient Q and remainder R such that, N = QS+R, R<N
Thus, based on the above definition of "successive division" we get,
N = 5a+1, where a = 6b+2, b = 7c+4 and c = 8d+5. Subsitituting these values back into the main equation we get,
N = 1680*d+1181
Pick any value for N now = 1181 (to simplify the calculations).
We get a remainder of 5 and quotient of 147 when divided by 8, remainder of 0 and quotient of 21 when divided by 7 , remainder of 3 and quotient of 3 when divided by 6 and remainder of 3 and quotient of 0 when divided by 5.
Thus, the sum of all the remainders = 5+0+3+3 = 11. D is the correct answer.