The positive integers mm and nn leave remainders of 2 and 3, respectively, when divided by 6. m>nm>n.
What is the remainder when m–nm–n is divided by 6?
Answer: C
We are given that the numbers mm and nn, when divided by 6, leave remainders of 2 and 3, respectively.
Hence, we can represent the numbers mm and nn as 6p+26p+2 and 6q+36q+3, respectively, where pp and qq are suitable integers.
Now,
m−n=(6p+2)−(6q+3)
=6p−6q−1
=6(p−q)−1
A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:
6(p−q)−1=6(p−q)−6+6−1
=6(p−q)−6+5
=6(p−q−1)+5
Thus, the remainder is 5