Find 3 consecutive even integers such that the sum of 3 times the first, 5 times the second, and 2 times the third is 418.

Answer 1

1st = x

2nd = x+2

3rd = x+4

3x + 5(x+2) + 2(x+4) = 418

3x + 5x + 10 + 2x + 8 = 418

10x + 18 = 418

10x = 400

x = 40

40, 42, 44 ---> 3*40 + 5*42 + 2*44 =

120 + 210 + 88 =

330 + 88 = 418