Quotient by Digit Sum

Let the number be $N=\overline{abc}$ . Then the value we want to maximize is $\frac{100a +10b + c}{a + b + c} = 1 + \frac{99a+9b}{a+b+c}$ . This is maximized when $c=0$ . Then we repeat again, to get $10+\frac{90a}{a+b}$ , which is maximized when $b=0$ . Then $M = 100$ , which holds for $N = 100, 200, \ldots, 900$ . So there are 9 such numbers.

Let $N=100a+10b+c$ , where $(1 \le a \le 9),(0 \le b,c \le 9)$ .

So $S(N)=a+b+c$

If $\frac{N}{S(N)}>100$ , then

$\frac{100a+10b+c}{a+b+c}>100$

$100a+10b+c>100(a+b+c)$

$100a+10b+c>100a+100b+100c$

$0>90b+99c$

$0>10b+11c$

This is impossible since $b$ and $c$ are non-negative.

If $\frac{N}{S(N)}=100$ , then it simplifies to

$0=10b+11c$

This is possible. Hence, $M=$ the maximum of $\frac{N}{S(N)}=100$ , and this occurs if $b=c=0$ .

$a$ is any number from $1$ to $9$ . So $N$ can be $100, 200,..., 900$ .

This gives us $9$ three-digit numbers which satisfy the required condition.

let D(N)=N/S(N),N=abc,so D(N)=(100a+10b+c)/(a+b+c)=100-(90b+99c)/(a+b+c),let b=c=0,we get the max. 100, this can be get when N is multi. of 100

Let $N = \overline{abc} = 100a + 10b + c$ , where $1 \leq a \leq 9$ , $0 \leq b \leq 9$ and $0 \leq c \leq 9$ .

We have $\frac {\overline{abc}} {a+b+c} = \frac{100a+10b+c}{a+b+c} = \frac {99a + 9b}{a+b+c} + 1$ . Thus, the maximum occurs when $c = 0$ .

Then $\frac {99a + 9b}{a+b} + 1 = \frac {90 a} {a+b} + 9 + 1$ . Thus, the maximum occurs when $b=0$ .

Then $\frac {90 a} {a} + 9 + 1 = 100$ would be the value of $M$ . There are $9$ possible values of $a$ , thus there are $9$ 3-digit numbers that satisfy $M = 100 = \frac {N}{S(N)}$ .