If $a, b$ and $c$ are positive integers such that $3a=4b=7c$, what is the smallest possible value of $a+b+c$?
(a) 84
(b) 36
(c) 61
(d) 56
(e) 42
The correct option is (c).

Let $N$ be the number given by $N=3a=4b=7c$. Then, the number $N$ is a multiple of 3, 4, and 7. Therefore, when we factorize the number $N$ into prime factors, at least the factors 2, 3, and 7 appear, with the first having an exponent of at least 2. It follows that $N$ is a multiple of $2^{2} \times 3 \times 7=84$. On the other hand, the numbers $a=4 \times 7=28$, $b=3 \times 7=21$, and $c=4 \times 3=12$ satisfy the equations $3a=4b=7c$. Therefore, $a=28$, $b=21$, and $c=12$ are the smallest possible values for $a$, $b$, and $c$, and $28+21+12=61$ is the smallest possible value for $a+b+c$.