$32 \cdot 12$ divided by 7 leaves a remainder of 5, divided by 5 leaves a remainder of 2, and divided by 3 leaves a remainder of 1. The sum of all three-digit numbers that satisfy these conditions is
(A) 2574.
(B) 3681.
(C) 4249.
(D) 4436.
(China Junior High School Mathematics Correspondence Competition, 1988)
[Solution] Let the number that satisfies the conditions "divided by 7 leaves a remainder of 5, divided by 5 leaves a remainder of 2, divided by 3 leaves a remainder of 1" be $N$, and let $x, y, z$ be the integer quotients when $N$ is divided by $7, 5, 3$ respectively. Then we have
$$
7 x+5=5 y+2=3 z+1=N,
$$

which means
$$
\left\{\begin{array}{l}
5 y-7 x=3 \\
3 z-5 y=1
\end{array}\right.
$$

From (2) we get
$$
y=\frac{7 x+3}{5}=x+\frac{2 x+3}{5} \text {. }
$$

Let $\frac{2 x+3}{5}=t$, then $y=x+t$,
and
$$
x=\frac{5 t-3}{2}=2 t-1+\frac{t-1}{2} \text {. }
$$

Let $\frac{t-1}{2}=n$, then $t=2 n+1$.
Also, $x=2 t-1+n=5 n+1, \quad y=x+t=7 n+2$,
From (3) we have $z=\frac{5 y+1}{3}=\frac{35 n+11}{3}=11 n+4+\frac{2 n-1}{3}$.
To make $z$ an integer, take $n=2$, then $z=27$. Substituting into (1), we get $N=82$.
Since the least common multiple of 7, 5, 3 is 105, the number that satisfies the problem can be expressed in the form $82+$ $105 k$ ( $k$ is an integer).
Actually, when $k=1,2,3,4,5,6,7,8$, $82+105 k$ is a three-digit number. Therefore, the sum we seek is
$$
82 \cdot 8+105 \cdot(1+2+3+4+5+6+7+8)=4436 \text {. }
$$

Thus, the answer is $(D)$.
Note: In ancient Chinese mathematical texts, such as "The Mathematical Classic of Sunzi," there is a problem: "There is an object whose number is unknown. When counted in threes, there is a remainder of two; when counted in fives, there is a remainder of three; when counted in sevens, there is a remainder of two. What is the number of the object?" To solve such "unknown number" problems, the Ming dynasty mathematician Cheng Dawei introduced a mnemonic in "General Source of Computational Methods":
Three people walking together, seventy is rare, five plum trees, one is sweet,
Seven children reunite, just half a month, subtract one hundred and five, and you will know.

Using this mnemonic to find $n$ in this problem, we have
$$
n=5 \cdot 15+2 \cdot 21+1 \cdot 70=187 .
$$

Its general form is $187+105 \mathrm{~m}$ ( $\mathrm{m}$ is an integer).