If $\frac{\left(\frac{a}{c}+\frac{a}{b}+1\right)}{\left(\frac{b}{a}+\frac{b}{c}+1\right)}=11$, where $a, b$, and $c$ are positive integers, the number of different ordered triples $(a, b, c)$ such that $a+2 b+c \leq 40$ is
(A) 33
(B) 37
(C) 40
(D) 42
(E) 45
Solution

Simplify the expression $\frac{\left(\frac{a}{c}+\frac{a}{b}+1\right)}{\left(\frac{b}{a}+\frac{b}{c}+1\right)}=11$ as follows:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-292.jpg?height=208&width=347&top_left_y=308&top_left_x=629)

$$
\begin{aligned}
& \frac{a c}{b c}=11 \\
& \frac{a}{b}=11 \quad(\text { since } c\{0) \\
& a=11 b
\end{aligned}
$$

By substitution, the condition $a+2 b+c £ 40$ becomes $13 b+c £ 40$.

Since $b$ and $c$ are positive integers, then $b$ can only take on the values 1,2 , or 3 . The values of $a$ correspond directly to the values of $b$, since $a=11 b$.

If $b=3$, there is one corresponding value of $c$. When $b=2$, there are 14 possible values of $c$. Finally if $b=1$, there are 27 possible values of $c$.

Therefore, the number of different ordered triples satisfying the given conditions is $1+14+27=42$.

ANSWER: (D)

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