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If you have 10 cubes ranging in size from 1cm through to 10cm cubed, is it possible to fit them in a rectangular box less than 4000cm cubed in two different combinations?
Hi Dean,
The sum of the cubes of the first n integers i.e
1³ + 2³ + 3³ + .... + n³ = [n(n+1)/2]²
For n = 10, the sum is
[10(11)/2]² = 55² = 3025
Depending on the arrangement it might or might not be possible for a box with less than 4000 cubic units of volume. Seems unlikely though but i only have time to do the maths for you.
Regards
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Answers by Expert:
Ahmed Salami
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I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I can as well help a good deal in Physics with most emphasis directed towards mechanics.
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An engineering graduate. I have been doing maths and physics all my life.
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