Determine the number of triples $(a, b, c)$ of three positive integers with $a<b<c$ whose sum is 100 and whose product is 18018 .
We note that $18018=18 \times 1001=2 \times 3^{2} \times 11 \times 91=2 \times 3^{2} \times 7 \times 11 \times 13$.

We want to find all triples $(a, b, c)$ of positive integers with $a+b+c=100$ and $a b c=18018$.

First, we note that none of $a, b, c$ can be a multiple of more than one of 7,11 or 13:

If one was a multiple of 11 and 13 , it would be at least 143 which is too large.

If one was a multiple of 7 and 13 , it would be at least 91 . It could not be a larger multiple of 91 , so the remaining two integers would have a product of $2 \times 3^{2} \times 11$ and a sum of 9 , which is not possible.

If one was a multiple of 7 and 11 , it would be at least 77 . It could not be a larger multiple of 77 , so the remaining two integers would have a product of $2 \times 3^{2} \times 13$ and a sum of 23 , which is not possible.

Therefore, suppose that $a=7 k, b=11 m$, and $c=13 n$ for some positive integers $k, m, n$. We ignore the restriction that $a<b<c$ for the moment and then incorporate this later on. Since $a b c=2 \times 3^{2} \times 7 \times 11 \times 13$, then $(7 k)(11 \mathrm{~m})(13 n)=2 \times 3^{2} \times 7 \times 11 \times 13$ or $k m n=18$. So we check the possible combinations of $k, m, n$ that have a product of 18 to see if the resulting sum $a+b+c$ equals 100 . We note that none of $k, m$ or $n$ can equal 18 (otherwise the corresponding value of $a, b$ or $c$ would be larger than 100):

| $k$ | $m$ | $n$ | $a$ | $b$ | $c$ | $a+b+c$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 9 | 7 | 22 | 117 | 146 |
| 1 | 9 | 2 | 7 | 99 | 26 | 132 |
| 2 | 1 | 9 | 14 | 11 | 117 | 142 |
| 2 | 9 | 1 | 14 | 99 | 13 | 126 |
| 9 | 1 | 2 | 63 | 11 | 26 | 100 |
| 9 | 2 | 1 | 63 | 22 | 13 | 98 |
| 1 | 3 | 6 | 7 | 33 | 78 | 118 |
| 1 | 6 | 3 | 7 | 66 | 39 | 112 |
| 3 | 1 | 6 | 21 | 11 | 78 | 110 |
| 3 | 6 | 1 | 21 | 66 | 13 | 100 |
| 6 | 1 | 3 | 42 | 11 | 39 | 92 |
| 6 | 3 | 1 | 42 | 33 | 13 | 88 |
| 2 | 3 | 3 | 14 | 33 | 39 | 86 |
| 3 | 2 | 3 | 21 | 22 | 39 | 82 |
| 3 | 3 | 2 | 21 | 33 | 26 | 80 |

Therefore, there are two triples of positive integers with the required properties.

These are $(a, b, c)=(11,26,63)$ and $(a, b, c)=(13,21,66)$. (We have rearranged each triple in increasing order.)

ANSWER: 2