Given $a+2 b+3 c+4 d+5 e=k$ and $5 a=4 b=3 c=2 d=e$, find the smallest positive integer value for $k$ so that $a, b, c, d$, and $e$ are all positive integers.
(A) 87
(B) 522
(C) 10
(D) 120
(E) 60

## Part C: Each question is worth 8 credits.
Given $a+2 b+3 c+4 d+5 e=k$ and $5 a=4 b=3 c=2 d=e$ find the smallest positive integer value for $k$ so that $a, b, c, d$, and $e$ are all positive integers.
(A) 87
(B) 522
(C) 10
(D) 120
(E) 60

## Solution

From the equalities $5 a=4 b=3 c=2 d=e$, we conclude that $e$ is the largest and $a$ is the smallest of the integers. Since $e$ is divisible by $5,4,3$, and 2 , the smallest possible value for $e$ is 60 . The corresponding values for $a, b, c$, and $d$, are $12,15,20$, and 30 , respectively. Thus, the smallest positive integer value for $k$ is $12+2(15)+3(20)+4(30)+5(60)=522$.

## Part C: