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I have to tell if the following is true or false:

$$(1234657)! +1 \equiv_{11111} (7654321)! -1$$

so by definition we can rewrite the previous equivalence as:

$(1 \cdot 2 \cdot \ldots \cdot 11110 \cdot 11112 \cdot \ldots \cdot 1234567) \cdot 11111 +1 \equiv_{11111} (1 \cdot 2 \cdot \ldots \cdot 11110 \cdot 11112 \cdot \ldots \cdot 7645321) \cdot 11111 -1$

that is

$a \cdot 11111 +1 \equiv_{11111} b \cdot 11111 - 1$

therefore

$0+1 \equiv_{11111} 0 - 1 $

hence however

$ 1 \not\equiv_{11111} 11110$

So the answer to the question is false. Is my reasoning correct?

wiskundeliefhebber
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haunted85
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  • Why a.11111 and b.11111 are zero ? – Grijesh Chauhan Jan 28 '13 at 17:01
  • @GrijeshChauhan because $a \cdot 11111$ and $b \cdot 11111$ are obviously multiple of $11111$ so if you divide $a \cdot 11111$ or $b \cdot 11111$ by $11111$ the resulting reminder is in both cases $0$. – haunted85 Jan 28 '13 at 17:07
  • ok . means remainder not multiple. I am not in practice on math but I like your that's what I asked – Grijesh Chauhan Jan 28 '13 at 17:10
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    right logic but wrong use of the linking word "hence", should be replaced by something like "but" or "however" which indicates there are something wrong in the argument before the linking word. – achille hui Jan 28 '13 at 17:16

1 Answers1

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Yes. It is perfect. $$ \begin{align} \end{align} $$

Berci
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