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The riddle: How can 8+9+10=7. Our idea is that 8,9,10 is in a number base different from the number base that the answer 7 is in. We have tried several bases but can't find the answer. Any other idea is also welcomed.

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closed as off-topic by T. Bongers, José Carlos Santos, Lord Shark the Unknown, Hans Lundmark, Magdiragdag Dec 20 '17 at 22:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – T. Bongers, José Carlos Santos, Lord Shark the Unknown, Hans Lundmark, Magdiragdag
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ If this is one of those "move a line to make this equation true" riddles then you could also move the 1 in front of the 10 to make $8+9+0=17$ $\endgroup$ – Tob Ernack Dec 20 '17 at 18:41
  • $\begingroup$ Is it worth noting that $8+9-10=7$ ? Or $\lfloor \frac{8 . 9}{10}\rfloor = 7$ $\endgroup$ – Gribouillis Dec 20 '17 at 18:43
  • $\begingroup$ Let's define $x\oplus y=\frac{x+y}{-1}$ then $8\oplus 9\oplus 10=7$ $\endgroup$ – zwim Dec 20 '17 at 22:02
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It could be if you take addition modulo $10$. And if we are trying to find all modulo that works we have to solve $27 \equiv _n 7$ we get $n\mid 20$ and $n>7$. So the only modulo that works are $10$ and $20$.

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    $\begingroup$ with modulo 20 too $\endgroup$ – zwim Dec 20 '17 at 18:46
  • $\begingroup$ Modulo $2$ and modulo $5$ also work. $\endgroup$ – Gribouillis Dec 20 '17 at 18:48

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