A sequence is arithmetic if the numbers increases by a fixed amount. For example, 2, 5, 8, are in an arithmetic sequence with a common difference of 3, and if these numbers represented the side lengths of a cuboid, the volume, V = 258 = 80 units3.
How many cuboids exist for which the volume is less than 100 units3 and the integer side lengths are in an arithmetic sequence?
Although 333 = 27 is a cuboid with a volume less than 100 units3, we shall discount cubes on the grounds that 3, 3, 3, is a trivial example of an arithmetic sequence.
By considering the first term, a, and the common difference, d, we have a method of systematically listing all of the solutions:
a=1, d=1, 123 = 6
a=1, d=2: 135 = 15
a=1, d=3: 147 = 28
a=1, d=4: 159 = 45
a=1, d=5: 1611 = 66
a=1, d=6: 1713 = 91
a=2, d=1: 234 = 24
a=2, d=2: 246 = 48
a=2, d=3: 258 = 80
a=3, d=1: 345 = 60
Hence there are exactly ten cuboids.
What is the maximum volume obtainable for a cuboid with side lengths in an arithmetic sequence and having a volume less than 1000 units3?
Can you generalise for a volume less than V units3?