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When I typed in

6^-1 mod 49 in Wolfram|Alpha, it gave me an answer of 41. Link here

If I type the same thing as (1/6) mod 49 , I don't see 41 any more. Why is this happening ?Link here

A Related question :

How is the answer to 6^-1 mod 49 , 41 in the first place ?

A small change in the above question :

If the question is find , 9^-2 mod 49, what are the steps to find the answer ?

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    $\begingroup$ $6\times41=246=245+1=(5)(49)+1\equiv1\pmod{49}$ $\endgroup$ – Gerry Myerson Jul 20 '12 at 9:23
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    $\begingroup$ You can use the extended Euclidean algorithm to find the modular inverse of $81$, $\bmod 49$. See this. $\endgroup$ – J. M. is a poor mathematician Jul 20 '12 at 9:35
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The $41$ is because $6\times 41=246=1\bmod 49$.

I assume that what's happening is that if you type $(1/6)$ instead, it assumes that you're happy to work with rational numbers rather than just integers and gives you back $1/6$.

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    $\begingroup$ In short: Mod[1/6, 49] and PowerMod[6, -1, 49] are two different things. $\endgroup$ – J. M. is a poor mathematician Jul 20 '12 at 9:25

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