The numbers a,b,c and d satisfy the equations: a+2b+3c+4d=k and 4a=3b=2c=d. What is the smallest value of k for which a,b,c and d are positive integers?

The key thing to notice here is that d must have the factors 4,3 and 2. The number 12 is the smallest to contain all those factors. We can then deduce c=6, b=4 and a=3 to get k=77.