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The numbers can be represented as n - 2, n and n + 2 so that (n - 2)n(n + 2) = 29667

This can be expressed as n(n2 - 4) = n3 - 4n = 39667 and the roots solved for a cubic equation.

However, the problem can also be solved by factoring.

If the numbers are a, b and c then a x b x c = 29667 and a, b and c are clearly factors of 29667.

The factors of 29667 are 3 x 11 x 29 x 31 which can be stated as 33 x 29 x 31.

The three consecutive numbers are therefore 29, 31 and 33.

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Q: What 3 consecutive odd numbers have a product of 29667?
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