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put this solution on YOUR website!find all sets of three consecutive multiples of 4 whose sum is between 72 and 108
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Let n = the smallest number
Then the other two numbers are n+4 and n+8
The sum is n + n+4 + n+8 = 3n + 12
72 <= 3n+12 <= 108
This gives
72 <= 3n+12
60 <= 3n
20 <= n
3n+12 <= 108
3n <= 96
n <= 32
So n ranges from 20 to 32. This gives the following sets:
20, 24, 28
24, 28, 32
28, 32, 36
32, 36, 40
