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Interesting question. You really should work on it yourself before you read on ....

Start by factoring 144 to see what we are dealing with: 144 = 12x12 = 4x3x4x3 = 2x2x2x2x3x3.

The lengths of the edges can be 2, 4, 6, 8, 9, 12, 16, 18, 24, and 36. (These are all the ways you can choose factors from {2,2,2,2,3,3} and still have 2 left over for the other 2 edges.)

Suppose one edge is 2. The other 2 edges could be 2 and 36, 3 and 24, 4 and 18, 6 and 12 or 8 and 9. This is already 5 "boxes".

Suppose one edge is 3 (but leave out 3,2,24 - we got that already). The other 2 edges could be 3 and 16, 4 and 12 or 6 and 8.

Suppose one edge is 4 (But leave out 3,4,12 - we just got that. Also we got 2,4,18 in the first step). The other edges could be 4 and 9 or 6 and 6.

That makes 10!

Answer:

2cm by 2cm by 36cm,

2cm by 3cm by 24cm,

2cm by 4cm by 18cm,

2cm by 6cm by 12cm,

2cm by 8cm by 9cm,

3cm by 3cm by 16cm,

3cm by 4cm by 12cm,

3cm by 6cm by 8cm,

4cm by 4cm by 9cm and

4cm by 6cm by 6cm.

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Q: Nelsons mathematics 8 identify 10 different rectangular prisms with whole number dimensions that have a volume of 144 cm 3?
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