Help!!!!!!!!!!!!!!!!!!!!?

Question

A fraction m/n. (m+1)/(n+1), (m+2)/(n+2), (m+3)/(n+3), (m+4)/(n+4) and (m+5)/(n+5) are reducible. I have got the answer 1/31. But I don't know how to get it. Can you do it?

Answer 1

Actually, there is a pattern here.
Let p be a prime congruent to 1 mod 3 and
also to 1 mod 5(therefore 1 mod 15). Let
the desired fraction be 1/p.
Then 2/(p+1), ... 6/(p+5) are all reducible.
To see this note that
p+1, p+3 and p+5 are all even and the numerators
of these fractions are even.
Finally, p+2 is divisible by 3, so 3/(p+2) is reducible.
and p+4 is divisible by 5, so 5/(p+4) is reducible.
Update: I don't think p needs to be prime here.
Any odd number congruent to 1(mod 30) will
work just as well.
Example: 361 = 19², works just fine!

Answer 2

This question does not really have a unique solution.

1/61 also works. So does 1/91, 1/121 etc.

You just have to recognize it I guess. I don't think there is any systematic method to finding those answers.