For each positive integer $n$, define $s(n)$ to be the sum of the digits of $n$. For example, $s(2014)=2+0+1+4=7$. Determine all positive integers $n$ with $1000 \leq n \leq 9999$ for which $\frac{n}{s(n)}=112$.
Consider a positive integer $n$ in the range $1000 \leq n \leq 9999$. Such an integer is of the form abcd for some digits $a, b, c, d$.

Note that $n=1000 a+100 b+10 c+d$, since the thousands digit is $a$, the hundreds digit is $b$, the tens digit is $c$, and the units digit is $d$. Also, $s(n)=a+b+c+d$.

Therefore, we want to find all $n=a b c d$ for which $\frac{1000 a+100 b+10 c+d}{a+b+c+d}=112$.

This is equivalent to $1000 a+100 b+10 c+d=112 a+112 b+112 c+112 d$ or $888 a-111 d=12 b+102 c$ or $296 a-37 d=4 b+34 c$ or $37(8 a-d)=4 b+34 c$.

Since the LS is a multiple of 37 , then the RS must also be a multiple of 37 .

Note that the RS and is also even, and so is an even multiple of 37 .

If $4 b+34 c=0(37)=0$, then $b=c=0$ and $8 a-d=0$. Since not all digits are zero, then we can have $a=1$ and $d=8$.

If $4 b+34 c=2(37)=74$, there are no values of $b$ and $c$ in the appropriate range that work. (This is because $c=0$ and $c=1$ would give values of $b$ larger than $9, c=2$ gives a value of $b$ that is not an integer, and $c \geq 3$ give values of $b$ that are negative.)

If $4 b+34 c=4(37)=148$, then $b=3$ and $c=4$ are the only values that work. This gives $8 a-d=4$. For $a$ and $d$ to be digits, we must have $a=1$ and $d=4$.

If $4 b+34 c=6(37)=222$, there are no values of $b$ and $c$ that work as in the case above.

If $4 b+34 c=8(37)=296$, then $b=6$ and $c=8$ are the only values that work. This gives $8 a-d=8$. For $a$ and $d$ to be digits, we must have $a=1$ and $d=0$ or $a=2$ and $d=8$. In total, there are four integers $n$ that satisfy the required: 1008, 1344, 1680, and 2688.

ANSWER: $1008,1344,1680,2688$