, 2024

Evaluating, we get $f(132)=132+1+3+2=138$. Suppose that $n$ is equal to the 3-digit positive integer $abc$. Then $f(n)=f(abc)=100a+10b+c+a+b+c=101a+11b+2c$. Since $f(n)=175$, then $101a+11b+2c=175$. It cannot be the case that $a $\geq$
2$, since if we had$a $\geq$ 2$, then$101a $\geq$ 202 which is too large, noting that

11b+2c$is always at least$0$. Therefore,$a < 2$which means that$a = 1$. When$a = 1$, we get$101+11b+2c=175$or$11b+2c=74$. It cannot be the case that$b $\geq$
7$, since if we had$b $\geq$ 7$, then$11b $\geq$ 77 which is too large, noting that

2c$is always at least$0$. Therefore,$b < 7$. If$b=6$, then$66+2c=74$or$2c=8$, and so$c=4$. If$$b\leq5$$, then$$11b\leq55$$, and so$$2c\geq74-55=19$$, which is not possible since$$c\leq9$$. We can confirm that$f(164)=164+1+6+4=175$, and so$n=164$. Suppose that$n$is equal to the 3-digit positive integer$pqr$. Then$f(pqr)=100p+10q+r+p+q+r$, and so$101p+11q+2r=204$. If$$p\geq3$$, then$$101p\geq303$$, and so$p=1$or$p=2$. If$p=1$, then$101+11q+2r=204$or$11q+2r=103$. Since$r $\leq$ 9$, then$2r $\leq$ 18$and so$11q $\geq$ 103 - 18 = 85$. Therefore,$q=8$or$q=9$. If$q=8$, then$88+2r=103$or$2r=15$, which is not possible since$r$is an integer. If$q=9$, then$99+2r=103$or$2r=4$, and so$r=2$. In this case,$n=192$and we can confirm that$f(192)=192+1+9+2=204$. If$p=2$, then$202+11q+2r=204$or$11q+2r=2. The only possible solution to

11q+2r=2$is$q=0$and$r=1$. In this case,$n=201$and we can confirm that$f(201)=201+2+0+1=204$. Therefore, if$f(n)=204, then the possible values of

n$are$192$and$201$.