Find Four consecutive integers such that 8 times the sum of first and third is 40 greater than 10 fourth?

Let the four consecutive integers be \(x, x + 1, x + 2, x + 3\).

Given that $$8(x + x + 2) = 40 + 10 ( x + 3) $$ $$ 8(x + x + 2) = 14 x + 30$$ $$ 16 x -14x = 30-16 $$ $$ 2 x = 14 $$ $$ x = 7 $$

Therefore, the four required consecutive numbers are 7, 8, 9 and 10.