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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.

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    Why would you multiply? What is your motivation? – 2017-01-29
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    Your first instinct is correct. Think in terms of what *even* numbers you *cannot* have, given that you do have a specified count of *at least* 26. – 2017-01-29
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    It cannot be that I have only 2 boxes such that the first box has 8 red balls and the second box had 18 red balls. So I thought of 'opening' a few more boxes. I thought of it this way: Suppose I opened two boxes, and both boxes have 8 red balls, then I would not have known that the boxes contained 18 red balls. Similarly, if I have 3 boxes, and if I opened all 3 boxes only to have 8 red balls in each box, then I wouldn't have realized that could be 18 red balls. So a guaranteed solution is to open a few boxes that have 8 or 18 red balls, hence, I was thinking of the lowest common multiple. – 2017-01-29
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    You opened *all* the boxes, and each box yielded one of those two results. If there were $x$ boxes holding $8$ red balls, and $y$ boxes holding $18$, (for some $x\geq 1, y\geq 1$) then there would be $8x+18y$ red balls. You only told us you have "a few boxes"; how many is that? – 2017-01-29
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    (Unless Jerry is revealing he has knowledge of the count of boxes through his statement....Ah...) – 2017-01-29

1 Answers 1

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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.

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    I am wondering if you meant gcd(a,b) = 1? Similarly, is it r = ab - a - b ∈ A? Because if I assume a = 9 and b = 4, then r = 23, and 23 ∉ A. – 2017-01-29
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    I edited the answer :) – 2017-01-29
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    Hi SSepehr, thank you for your help! I'm just wondering, why did you want to use values of a and b such that gcd(a,b) = 1? In addition, how did you happen to conclude that the answer is 2r? Why 2? Am I right to say that the "2" came from the gcd(18, 8)? – 2017-01-30
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    @Icycarus Hi. Yes, the set we're looking for is $B = \{8m+18n|m, n \ge 0\}$, so everything is doubled at the end. – 2017-01-30
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    Ahh! I got it! Thank you! I have gotten the answer already! – 2017-01-30