Number to digit sum ratio

It's easy to write some code for this, but in case anyone is interested in a more manual approach:

For a four-digit number $n=\overline{abcd}$ , the ratio we're interested in is $r=\frac{1000a+100b+10c+d}{a+b+c+d}$ . We can find out information about the digits that minimise $r$ with repeated division.

$r = \frac{1000a+100b+10c+d}{a+b+c+d} = \frac{999a+99b+9c+a+b+c+d}{a+b+c+d}$

$r = 1+\color{#D61F06}\frac{999a+99b+9c}{a+b+c+d}$

The numerator of the red fraction doesn't depend on $d$ ; so we can minimise it by making $d$ as large as possible, ie $\boxed{d=9}$ .

Now $r=1+\frac{999a+99b+9c}{a+b+c+9}$ . Using the same trick, we have

$r=1+\frac{999a+99b+9c}{a+b+c+9}=1+\frac{990a+90b+9(a+b+c+9)-81}{a+b+c+9}$

$r = 10+\color{#D61F06}\frac{990a+90b-81}{a+b+c+9}$

By the same logic as before, the red fraction (which is positive, since $a \ge 1$ ) is minimised by taking $\boxed{c=9}$ .

Once again: $r = 10+\frac{990a+90b-81}{a+b+18}$

$r = 10+\frac{900a+90(a+b+18)-1701}{a+b+18}$

$r = 100+\color{#D61F06}\frac{900a-1701}{a+b+18}$

Now we can make the red fraction negative by taking $a=1$ . This is a much bigger effect than we get by changing $b$ ; so we take $\boxed{a=1}$ .

Finally, we have $r = 100-\frac{801}{b+19}$ . This is clearly minimised by taking $\boxed{b=0}$ ; so the ratio is minimised by taking $n=1099$ giving the answer $\boxed{\frac{1099}{19}}$