Find four consecutive even integers such that four times the greatest is 36 more than more than the sum of the other three.Four times a number increased by 5 is 37.Find the number.

Answer 1

If " n " is an integer, then n, n + 2, n + 4 and n + 6 will be even consecutive integers.

Four times the greatest is 36 more than more than the sum of the other three.

This mean :

4 ( n + 6 ) = 36 + ( n + n + 2 + n + 4 )

4 ( n + 6 ) = 36 + ( n + n + 2 + n + 4 )

4 * n + 4 * 6 = 36 + 3 n + 2 + 4

4 n + 24 = 42 + 3 n Subtract 3 n to both sides

4 n + 24 - 3 n = 42 + 3 n - 3 n

n + 24 = 42 Subtract 24 to both sides

n + 24 - 24 = 42 - 24

n = 18

So consecutive even integers are :

18

18 + 2 = 20

18 + 4 = 22

18 + 6 = 24

Proof :

4 * 24 = 36 + 18 + 20 + 22

96 = 96