Summing Ratio of Integers Over Their Digital Sums

We divide the three digit numbers into 6 groups, depending on how many zeroes and repeated digits they have.

Case 1. No zeroes, no repeated digits.

$\frac{ \overline{abc} } { a+b+c } + \frac{ \overline{acb} } { a+b+c } + \frac{ \overline{bca} } { a+b+c } + \frac{ \overline{bac} } { a+b+c } + \frac{ \overline{cab} } { a+b+c } + \frac{ \overline{cba} } { a+b+c } = \frac{ 222 ( a + b + c ) } { a+b+ c } = 222$

There are ${ 9 \choose 3 }$ possible ways to choose these numbers, so their sum is ${ 9 \choose 3 } \times 222 = 18648$ .

Case 2: No zeroes, only 2 repeated digits

$\frac{ \overline{aab} } { a+a+b } + \frac{ \overline{aba} } { a+a+b } + \frac{ \overline{baa} } { a+a+b } = \frac{ 111 ( 2a + b ) } { 2a + b } = 111$

There are $9 \times 8$ possible ways to choose these numbers, so their sum if $9 \times 8 \times 111 = 7992$ .

Case 3: No zeroes, all 3 repeated digits.

$\frac{ \overline{aaa} } { a + a + a } = \frac{ 111 \times a } { 3a } = 37$

Since there are 9 possible ways, their sum is $9 \times 37 = 333$ .

Case 4: 1 zero, two distinct non-zero digit

$\frac{ \overline{ab0} } { a+b+0 } + \frac{ \overline{a0b} } { a+b+0 } + \frac{ \overline{ba0} } { a+b+0 } +\frac{ \overline{b0a} } { a+b+0 } = \frac{ 211 ( a + b) } { a+ b } = 211$

Since there are ${ 9 \choose 2 }$ ways to choose these numbers, so their sum is ${ 9 \choose 2 } \times 277 = 7596$ .

Case 5: 1 zero, two repeated non-zero digit

$\frac{ \overline{aa0} } { a+a+0 } + \frac{ \overline{a0a} } { a+a+0 } = \frac{ 211 a } { 2a } = 105.5$

There are 9 ways to choose these numbers, so their sum is $9 \times 105.5 = 949.5$ .

Case 6: two zeros, one non-zero digit

$\frac{ \overline{a00}} { a + 0 + 0 } = 100$

There are 9 ways to choose these numbers, so their sum is $9 \times 100 = 900$ .

Summing up these 6 cases, we find that the sum is 36418.5

Ok, So there are two types of solution. One is discussed above already by Mahan Farzan.

I'd like to discuss another method that takes a bit of thinking but make our lives much easier.

We can rewrite the above equation as

$\left \{\sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{100a+10b+c}{a+b+c} \right ) -f_{3}(0,0,0) \right \} - \left \{ \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{10b+c}{b+c}\right ) -f_{2}(0,0)\right \}$

Here $f_{3} (0,0,0)$ is the value when $a=0, b=0, c=0$ . For that we can use the limit to find the value of $\frac{100a+10b+c}{a+b+c} = \frac{111a}{3a} = 37$ .
Similarly $f_{2}(0,0) = \frac{11b}{2b} = 5.5$ .

Now coming to the main part, The triple summation. It's clearly symmetric in $a, b$ and $c$ . Thus there are 6 other similar permutations. Adding them together,

$6\sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{100a+10b+c}{a+b+c} \right ) = \sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{222a+222b+222c}{a+b+c} \right ) = \sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} 222 = 222000$

Thus, $\sum_{a=0}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{100a+10b+c}{a+b+c} \right ) = 37000$ .

Similarly, $2\sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{10b+c}{b+c}\right ) = \sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{11b+11c}{b+c}\right ) = \sum_{b=0}^{9} \sum_{c=0}^{9}11 = 1100$

Thus, $\sum_{b=0}^{9} \sum_{c=0}^{9} \left(\frac{10b+c}{b+c}\right ) = 550$ .

Thus The net sum is, $\left\{37000-37\right\} - \left\{550-5.5\right\} = 36450 - 31.5 = 36418.5$ .

Can we figure out a generalized solution for this?