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Arithmetic Volume

Problem

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 = 2times5times8 = 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?


Solution

Although 3times3times3 = 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,   1times2times3 = 6
a=1, d=2:   1times3times5 = 15
a=1, d=3:   1times4times7 = 28
a=1, d=4:   1times5times9 = 45
a=1, d=5:   1times6times11 = 66
a=1, d=6:   1times7times13 = 91
a=2, d=1:   2times3times4 = 24
a=2, d=2:   2times4times6 = 48
a=2, d=3:   2times5times8 = 80
a=3, d=1:   3times4times5 = 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?

Problem ID: 198 (10 Jan 2005)     Difficulty: 1 Star

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