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Found this problem in my SAT book the other day and wanted to see if anyone could help me out.

A positive integer is said to be "tri-factorable" if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable?

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You need to find the biggest tri-factorable integer less than $1000$ (recall that a tri-factorable integer can be written $n(n+1)(n+2)$ and this is roughly equal to $n^3$). –  Joel Cohen Aug 6 '12 at 20:32
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2 Answers

For numbers of this size, you can make a list: $1\times 2\times 3=6, 2\times 3 \times 4= \ldots$ and see where it gets larger than $1000$. A spreadsheet makes it quite easy.

If the upper limit were enough higher, you could let $n$ be the middle number. Then your trifactorable number is $(n-1)n(n+1)=n^3-n$ You can take the cube root of the upper limit and check whether you need one more.

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Awesome, thanks everyone! –  l34p3r Aug 7 '12 at 7:07
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$1,000= 10*10*10<10*11*12$ so in the product $n*(n+1)*(n+2)$, $n$ must be less then $10.$

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And as an extra hint $9 \times 10 \times 11 = 990 \lt 1000$ –  Henry Aug 6 '12 at 21:03
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