Find four consecutive integers such that when the difference of the greatest and least is combined with the sum of the middle two integers the result is 46.

Answer 1

We denote four consecutive integers as n, (n+1), (n+2) and (n+3). As stated in the problem

[(n+3) - n] + (n + 1) + (n + 2) = 46.

We simplify

n + 3 - n + n + 1 + n + 2 = 46; 2n + 6 = 46; 2n = 40; n = 20. Hence the integers are 20, 21, 22 and 23.

We check that (23 - 20) + 21 + 22 = 46. Hence our solution is correct.