If the equation 9x+13y=k has exactly 5 solutions, where x and y are positive integers, what is the minimum possible value of k ?

The answer is 490.
The LCM of 9 and 13 is 117.
If we are looking at non negative solutions (includes 0), the answer would be (5-1)*117=468. (Explanation for this given below)
As we are looking at positive integers and x and y have to be atleast 1, the answer is 468+9+13=490

As a brief answer, let solutions(k) be the number of non negative solutions to the equation 9x+13y=k
If k<117, solutions(k) is either 0 or 1.
If 117<=k<234, solutions(k) is either 1 or 2 and so on.
So, the least value of k for which solutions(k) is 5 is (5-1)*117.

For the purposes of CAT, remember this as a theorem. If px+qy=k has exactly 'n' non negative integral solutions, then k lies between (n-1)*LCM(p,q) and n*LCM(p,q)

Hi
Could you please explain this : As we are looking at positive integers and x and y have to be atleast 1, the answer is 468+9+13=490As we are looking at positive integers and x and y have to be atleast 1, the answer is 468+9+13=490

it's awesome, thank you very much for the solution

Let 9x+13y=k
Then y=(k-9x)/13
y=k/13-9x/13
As y is a positive integer for that x must be multiple of 13 and y has 5 values
for that x=13,26,39,52,65
Then minimum value of y=k/13-65*9/13=k/13-45
The minimum possible value of y=1
Then we can say 1=k/13-45
k/13=46
Or K=13*46=598