-2
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1 * 1 = 1

2 * 13 = 2

3 * 3 = 3

44 * 8 = 4

73 * 5 = 0

90 * 60 = 0


Q1) What is the smallest set of numbers under this operation that can equal 5? 6?

Q2) What is 8 * 9?

Hint:

The operation is commutative, but not associative.

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  • $\begingroup$ Can you clarify what Q1 means, please? It's a binary operation but the phrasing of the question makes it seem as though you can interpret it as a single-valued function. Is that right? $\endgroup$ – hexomino Jun 3 '16 at 12:35
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To meet the commutativity requirement I can make the answers to Q1

$0$ and $0$

using

$$L*R=\min(L\pmod{20}, R\pmod{20})\pmod{5}$$

in which case $8*9$ is

$3$

Note that the function is not associative, for example $(1*5)*5 \neq 1*(5*5)$


Previous answer:

I can make the answers to Q1

$0$ and $0$

using

$$L*R = \begin{cases} L\pmod5 & \text{if $L\pmod{5} \in \{1,2,4\}$} \\ R\pmod5 & \text{if $L\pmod{5} \in \{0,3\}$} \\ \end{cases}$$

in which case $8*9$ is

$4$

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  • $\begingroup$ It does have something to do with modulo 5, but not to do with the numbers on the left or right hand side. And 8*9 is not equal to 4. $\endgroup$ – Inazuma Jun 2 '16 at 7:16
  • $\begingroup$ The question asked to make the number of answers that are $5$ and then $6$ as small as possible and then to give $8*9$, this makes both as small as possible and then evaluates $8*9$ using the function I provided, so why is it not correct? $\endgroup$ – Jonathan Allan Jun 2 '16 at 7:25
  • $\begingroup$ I am looking for a specific operation. $\endgroup$ – Inazuma Jun 2 '16 at 7:30
  • $\begingroup$ Then you need to reword your puzzle - I see you now added a hint which rules out this answer, but does it rule out all possible answers except the one you have in mind? $\endgroup$ – Jonathan Allan Jun 2 '16 at 7:31
  • $\begingroup$ I find this no different than a riddle, which may also have possible answers. $\endgroup$ – Inazuma Jun 2 '16 at 7:32

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