2008 iTest Problems/Problem 30

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Problem

Find the number of ordered triplets $(a,b,c)$ of positive integers such that $a<b<c$ and $abc=2008$.

Solution

The prime factorization of 2008 is $2^3 \cdot 251$. Once again, use casework to organize the possibilities.

  • If $a = 1$, the number of possibilities of $b$ and $c$ are really the number of distinct pairs of two numbers that multiply to $2008$ that do not have $1$. In this case, there are $\tfrac{4 \cdot 2}{2} - 1 = 3$ possibilities.
  • If $a = 2$, then the only possibility that works is when $b = 4$ and $c = 251$, for another $1$ possibility.
  • If $a = 4$ (or higher), then there are no possibilities that work because one of the values of $b$ and $c$ would be smaller than $a$.

In total, there are $\boxed{4}$ ordered triplets that satisfy the conditions.

See Also

2008 iTest (Problems)
Preceded by:
Problem 29
Followed by:
Problem 31
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