When the smaller of two consecutive integers is added to five times the larger the result is 41?

Suppose n is the smaller integer. Then the larger integer is n + 1, so that 5 times the larger is 5*(n + 1).

Their sum is n + 5*(n + 1) = 6n + 5

Therefore 6n + 5 = 41

6n = 41 - 5 = 36

So that n = 36/6 = 6