There are four unequal, positive integers $a, b, c$, and $N$ such that $N=5 a+3 b+5 c$. It is also true that $N=4 a+5 b+4 c$ and $N$ is between 131 and 150 . What is the value of $a+b+c$ ?
(A) 13
(B) 17
(C) 22
(D) 33
(E) 36
There are four unequal, positive integers $a, b, c$, and $N$ such that $N=5 a+3 b+5 c$. It is also true that $N=4 a+5 b+4 c$ and $N$ is between 131 and 150 . What is the value of $a+b+c$ ?
(A) 13
(B) 17
(C) 22
(D) 33
(E) 36

## Solution

We are told that $N=5 a+3 b+5 c$ (1) and $N=4 a+5 b+4 c$ (2). Multiply equation (1) by 4 to get $4 N=20 a+12 b+20 c$ (3). Similarly, multiply equation (2) by 5 to get $5 N=20 a+25 b+20 c$ (4). Subtract equation (3) from equation (4) to get $N=13 b$.

Since $N$ and $b$ are both positive integers with $131<N<150, N$ must be a multiple of 13 . The only possible value for $N$ is 143 , when $b=11$.

Substitute $N=143$ and $b=11$ into equation (1) to get

$$
\begin{aligned}
143 & =5 a+3(11)+5 c \\
110 & =5 a+5 c \\
22 & =a+c
\end{aligned}
$$

Thus, the value of $a+b+c$ is $22+11=33$.

ANSWER: (D)