I came across this "problem". The problem goes like this: I bought a few boxes of red balls. My friend, Jerry, is trying to guess the number of red balls I have. When I opened up the boxes (not in front of Jerry), I realized that the boxes contained either 8 red balls or 18 red balls. I told this to Jerry. Jerry then immediately claimed that I could have any even number of red balls greater than x. What is x and prove that Jerry is correct.
My intuition is this: Isn't the answer just 8+18 = 26 red balls? Is the answer that straightforward? Or is there another logical answer for this 'problem'? I have another answer, which is 72 red balls (Because it's the lowest common multiple of 8 and 18).
I have no idea what is the correct answer to this question. Any suggestions would be great! Thank you!
P.S. I'm in a college so I don't think this is a 1st grade type of Mathematics problem.
Hint:
Let $a = \frac{18}{2} = 9$ and $b = \frac{8}{2} = 4$. Now $\gcd(a,b) = 1$, and consider $$A = \{am+bn|m, n \ge 0\}.$$ Now try to show that
$1)r = (a-1)(b-1) \in A$
$2)(a-1)(b-1)-1 \notin A$
$3)\forall k>r \space \space k \in A$.
Then the answer would be $2r$.
Note:
Number $26$ is possible, however $28$ isn't, and we're looking the smallest number $n$ for which all $k\ge n$ are possible.