Integer Triples Making 100

I used the $a$ just to fill in the difference between 100 and $b\times c$ . For every pair of integers $b\times c<100$ , there will be one and only one triple. I went from 1 to 100 for $b$ and for each value of $b$ , there are $\left \lfloor { \frac {99} {b} } \right \rfloor -1 +1$ possible values for $c$ . Summing this up is a little time consuming from 1 to 19, then the results become easier to calculate, because they come in large groups.

For $b=1$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 99$ . For $b=2$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=49$ . For $b=3$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 33$ . For $b=4$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =24$ . For $b=5$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 19$ . For $b=6$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 16$ . For $b=7$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=14$ . For $b=8$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 12$ . For $b=9$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 11$ . For $b=10, 11$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 9$ . For $b= 12$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=8$ . For $b=13,14$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 7$ . For $b=15, 16$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=6$ . For $b=17$ to $19$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 5$ . For $b=20$ to $24$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 4$ . For $b=25$ to $33$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =3$ . For $b=34$ to $49$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=1$ . For $b=50$ to $99$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =1$ .

Hence, the total number of positive integer solutions is $99 +49+33+24+19+16+14$ $+12+11+9\times 2 +8+7\times 2+6\times 2+5\times 3+4\times 5+3\times 9+ 2\times 16+1\times 50 =473$ .

[Details filled in - Calvin]

Since $a$ is at least 1, $b$ can take on any value from 1 to 99. For each value of $b$ , $c$ can be an integer from 1 to $\frac {99}{b}$ , and then we set $a = 100 - b \times c$ , to obtain a positive integer. As such, there are $\left \lfloor { \frac {99} {b} } \right \rfloor -1 +1$ possible values for $c$ . (Note: $\left \lfloor x \right \rfloor$ denotes the largest integer less than or equal to $x$ ). We now calculate $\displaystyle \sum_{b=1}^{99} \left \lfloor \frac {99}{b} \right \rfloor$ .

For $b=1$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 99$ . For $b=2$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=49$ . For $b=3$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 33$ . For $b=4$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =24$ . For $b=5$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 19$ . For $b=6$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 16$ . For $b=7$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=14$ . For $b=8$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 12$ . For $b=9$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 11$ . For $b=10, 11$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 9$ . For $b= 12$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=8$ . For $b=13,14$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 7$ . For $b=15, 16$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=6$ . For $b=17$ to $19$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 5$ . For $b=20$ to $24$ , we get $\left \lfloor \frac {99}{b} \right \rfloor = 4$ . For $b=25$ to $33$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =3$ . For $b=34$ to $49$ , we get $\left \lfloor \frac {99}{b} \right \rfloor=2$ . For $b=50$ to $99$ , we get $\left \lfloor \frac {99}{b} \right \rfloor =1$ .

Hence, the total number of positive integer solutions is $99 +49+33+24+19+16+14$ $+12+11+9\times 2 +8+7\times 2+6\times 2+5\times 3+4\times 5+3\times 9+ 2\times 16+1\times 50 =473$ .

Note: An alternative solution proceeds by from considering $a =1$ to $99$ and observing that there are $\sigma_0 (100-a)$ solutions. However, this requires listing all values of $\sigma_0 (n)$ , which is tedious. In fact, this method of double-counting shows that $\displaystyle \sum_{i=1}^N \sigma_0 (i) = \sum_{i=1}^N \left \lfloor \frac {N} {i} \right \rfloor$ .