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I need to find three integer $ a\equiv 19^{10562}(mod\: 40320) $. I know that I need to use Eulers Formula $a^{\phi (m)}\equiv 1(mod\: m)$. I have determined that $\phi (m)=9216$. So using this I think $19^{9216}\equiv 1(mod\, 40320)$. Now I believe that gets me to the following.

$$19^{10562}\equiv 19^{9216}\cdot 19^{1346}(mod\, 40320)$$

But after this I am unsure where to go with this because I'm not sure how to address the $19^{1346}(mod\, 40320)$. Any insight would be appreciated.

Thanks,

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    I don't know if I am correct, but I posted an answer. – 2017-02-18
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    Wait.... why the author deleted the question content? Can any moderator return it please? – 2017-05-26

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I don't know if I am correct, but let's see... Supposing I am correct, this is very easy.

Notice that $19^{10562}$ itself would be congruent to itself in mod 40320. So will $19^{10562}+40320$, and so will all numbers $19^{10562}+40320n$ where $n$ is an integer...

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    This is a fine answer to the question as written. I suspect that the poser intended that the smallest one be in the range $[0,40319]$ but why do the work? If you need an explicit base $10$ representation, [Alpha](http://www.wolframalpha.com/input/?i=19%5E10562) is happy to oblige if you click on More Digits enough. +1 – 2017-02-18
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    Sneaky, but accurate +1. I have written a solution to get low integers, but as Ross says, that wasn't asked for.... – 2017-02-18
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The Carmichael function $\lambda(40320)$ $=\lambda(2^7\cdot 3^2\cdot 5\cdot 7)$ $={\rm lcm}(\lambda(2^7), \lambda(3^2), \lambda(5), \lambda(7)) $ $={\rm lcm}(32, 6, 4, 6) =96$. So the value of $19^k\bmod 40320$ will cycle on a loop of length that divides $96$ and since $19$ is coprime to $40320$, we will have $19^{96}\equiv 1 \bmod 40320$.

Therefore $19^{10562} \equiv 19^{\large 10562 \bmod 96} \equiv 19^2\equiv 361 \bmod 40320.$ We can take e.g $k=0,1,2$ in $361+40320k$ to get three values as required.

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    Modulo arithmetic...the only branch of mathematics in which you can be off by $19^{10562}-361$ and still get the right answer. I waited myself to see an answer with integers closest to 0 as possible. +1 – 2017-02-18
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    It's perfectly valid to evaluate $19^{1346} \bmod 40320$ also to get the correct answer. $2944$ isn't a valid possibility for any $19^k \bmod 40320$ so something else is wrong in your calculations. – 2017-02-26
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    OK so far, maybe not the easiest way but OK. However you do know that $19^{\large(1024+256+64+2)}$ $= 19^{\large 1024}\cdot 19^{\large 256}\cdot 19^{\large 64}\cdot 19^{\large 2}$, right? – 2017-02-26
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    I would take the modulo after every multiplication, but otherwise yes. What values do you have for those four exponents? – 2017-02-26
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    There's your problem - you should have $19^{1024}\equiv 19^{256}\equiv 19^{64} \equiv 34561$ – 2017-02-26
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    $34561^2 = 1194462721$, you seem to be running out of precision. – 2017-02-26