118
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27 + 28 + 29 + 30 = 114
Three consecutive integers whose sum is 117 are 38, 39, and 40. N + (N+1) + (N+2) = 117 3N + 3 = 117 3N = 114 N = 38
The integers are 37, 38 and 39.
Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40. Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114 So 3n+6=114 and 3n=108 so n=36 then next two numbers must be 38 and 40 since they are consecutive even integers.
27, 28, 29 and 30