Question

Find three consecutive even integers whose sum is 66. show work pls

Answer 1

three consecutive even integers are x, x+2 and x+4

x + x+2 + x+4 = 66
3x + 6 = 66
3x = 66-6
3x = 60
x = 60/3
x = 20  ← first integer

x + 2 = 20 + 2 = 22 ← second  integer

x + 4 = 20 + 4 = 24 ← third integer

Answer 2

For our three integers to be even x%2=0 ∀x∈S where S= {our three integers}. To guarantee this set our first integer as 2n (ie. 2n%n≡0). Then to have three consecutive even integers increment from the initial value by 2 thus setting them as 2n, 2n+2 and 2n+4.

Thus:
$2n+2n+2+2n+4=66$

Simplifying:
$6n+6=66$

Solving for 'n':
$6n=60$
$n=10$

The integers are thus 2n=20, 2n+2=22 and 2n+4=24.

To check: 20+22+24=66.