Question 831136
Find four consecutive odd integers such that the sum of the second and the fourth is 63 more than one-fifth of the third?

Let 4 consecutive odd integers are x , x+2, x+4 , x+ 6

(x+2)+(x+6) = 63 + (1/5)(x+4)
2x + 8 = 63 + (1/5)(x+4)
2x - 55 = (1/5)(x+4)
10x - 275 = x+ 4
9x = 279
x = 31

So, consecutive odd integers are 31,33,35,37   Answer