Question 826231
Let the three integers be n1,n2 and n3 such that
n1 = n
n2 = n+1
n3 = n+2
We are told that the sum of twice the second and three times the third is five less than six times the first.
putting that into numbers,
2*n2 + 3*n3 = 6*n1 - 5
i.e. 2(n+1) + 3(n+2) = 6n - 5
2n + 2 + 3n + 6 = 6n - 5
rearranging the eqn gives us,
13 = n
n = 13
Answer: the three numbers are: 13, 14, 15