Question

One number is 6 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 412, find the numbers.

Answer 1

Given

n, 6n, n+100 sum to 412

Find

n, 6n, n+100

Solution

... n + 6n + (n+100) = 412

... 8n = 312 . . . . . . . . . . . . . collect terms, subtract 100

... n = 39 . . . . . . . . . . . . . . . divide by 8

... 6n = 234

... n+100 = 139

The numbers are 39, 234, and 139.