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9 years, 3 months ago
9 years, 3 months ago
Let's break this into cases with respect to common differences.
Common Difference = 0 => (1,1,1), (2,2,2), (3,3,3), (4,4,4)
Common Difference = 1 => (1,2,3), (2,3,4), (3,4,5)
Common Difference = 2 => (1,3,5), (2,4,6)
Common Difference = 3 => (1,4,7), (2,5,8)
Common Difference = 4 => (1,5,9)
Common Difference = 5 => (1,6,11)
Common Difference = 6 => (1,7,13)
Hence there are 14 possibilities such that the volume is less than 100 and the side lengths are in arithmetic progression.
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