Find three consecutive even integers such that the sum of the first, five times the second and four times the third is 1226.

Question

Find three consecutive even integers such that the sum of the first, five times the second and four times the third is 1226.

Answer 1

Dylan,

 

Let   n      = the first even integer

       n+2  = next even one

       n +4 = the third consecutive even integer

 

Your expression becomes   n + 5(n+2)  + 4(n+4) = 1226 , simplifying

10n + 26 = 1226 , and 

n = 1200/10 = 120 , so the first integer is indeed even

 

Thus, the 3 consecutive even integers are 120, 122, and 124.