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$A,B$ be subsets of $S=${$ 0,1,...,9$}, $ a\in A $ and $b \in B$ choosen randomly and $a+b$ takes each value $0,1,...,9$ with equal probalibility. Then it implies that one of $A$ or $ B$ is singleton.(It's a 'True or False' Question)

What I tried to find if there is any pair of set $A$ and $B$ which are not singleton.

As each of sum value $0$ to $9$ has equal probabilty. only to find zero is $0+0$ so $0$ appears in both $A$ and $B$. And for this reason we will get a unique choice of $a \in A$ and $b \in B$ such that $a+b=n$ for each $n \in S$.

In this way I am trying to find a pair of $A$ and $B$ which are not singleton. But i this will take so much time to check whether such pair exists. So can any one give any suggestion or hint!

I added the Actual Statement from the source:

"Let $A$, $B$ be subsets of {$0, . . . , 9$}. It is given that, on choosing elements $a\in A$ and $b \in B$ at random, $a + b$ takes each of the values $0, . . . , 9$ with equal probability. Then one of $A$ or $B$ is singleton"

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  • $\begingroup$ Are you assuming that both $A,B$ are non-empty? You didn't state that as an assumption, though you vaguely suggest it in the header. And did you mean to restrict $a+b$ to $\{0,\cdots, 9\}$? After all, in principle the sum could be as big as $18$. $\endgroup$
    – lulu
    Commented Oct 17 at 13:17
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    $\begingroup$ For example, the sets $A=\{6,7\}$, $B=\{7,8\}$ pass your tests. $a+b$ can not take any of the values in $\{0, \cdots, 9\}$ so it takes all of them with the same probability, namely $0$. Note that you didn't say anything about the probability that $a+b\in \{10, \cdots, 18\}$. $\endgroup$
    – lulu
    Commented Oct 17 at 13:20
  • $\begingroup$ Please edit to state all your assumptions clearly. Perhaps it would help to link to the source of the problem, so we could see the original wording. Failing that, please reproduce it without changing any of the words. $\endgroup$
    – lulu
    Commented Oct 17 at 13:25
  • $\begingroup$ @lulu Now you can see edited version $\endgroup$
    – PhilM
    Commented Oct 17 at 13:33
  • $\begingroup$ As mentioned, the statement as written has "degenerate" counterexamples. If $A=\emptyset$, say, then you can't choose $a,b$ as requested so each probability is $0$, which is the same for all the values. But I think my proposed counterexample (below) holds even if the problem statement were cleaned up. $\endgroup$
    – lulu
    Commented Oct 17 at 13:34

1 Answer 1

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As mentioned in the comments, the assumptions are very vague and unclear.

That said, I think that $$A=\{0,1\}\quad \&\quad B=\{0,2,4,6,8\}$$ is a counterexample, regardless of the details. (Taking $a=0$ gives you all the even numbers exactly once and taking $a=1$ gives you all the odd numbers exactly once).

Should add: I expect the problem intended to ask "Given two non-empty subsets $A,B$ of $\{0, \cdots, 9\}$ such that

I: the sum $a+b$ for $a\in A$ and $b\in B$ is always between $0$ and $9$ (inclusive) and

II: if $a,b$ are chosen uniformly at random and independently of each other then the values in $\{0, \cdots, 9\}$ are all hit with equal probability

then is it true that one of $A,B$ must be a singleton?" Certainly, my counterexample applies to that interpretation.

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  • $\begingroup$ Yes! this counter examples works $\endgroup$
    – PhilM
    Commented Oct 17 at 13:34
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    $\begingroup$ Under the assumption that the random process of choosing a and b independently at random from A and B respectively gives a 1/10 chance of hitting each element of S, that is, sums of more than 9 are impossible, the only other solution (up to permuting A and B) is A={0,5}, B={0,1,2,3,4}. $\endgroup$
    – F.U.A.S.
    Commented Oct 17 at 13:38

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