Once again 2016

201 6 201 5 201 4 201 3 . . . . 3 2 1 \large 2016^{2015^{2014^{2013^{..^{..^{3^{2^{1}}}}}}}}

Find the 2016th rightmost digit of the decimal representation of the number above.


You can use a calculator or computer program.


The answer is 4.

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2 solutions

Mr X
Dec 12, 2016

We proved in the solution of Once again 2016 simplified that 201 6 201 5 2014 S 2015 2016^{2015^{2014}}\in S_{2015} . Here we complete the solution proving that 201 6 201 5 201 4 . . . 3 2 1 S 2016 2016^{2015^{2014^{...^{3^{2^1}}}}} \in S_{2016} .

Note that 201 6 201 5 2014 S 2015 ( 201 6 201 5 2014 ) 5 = 201 6 5 201 5 2014 S 2016 2016^{2015^{2014}} \in S_{2015} \Rightarrow (2016^{2015^{2014}})^5=2016^{5*2015^{2014}}\in S_{2016} This means we only need to prove that 201 5 201 4 . . . 3 2 1 is a multiple of ( 5 201 5 2014 ) 2015^{2014^{...^{3^{2^1}}}}\text{ is a multiple of } (5*2015^{2014})

\Leftrightarrow

201 5 ( 201 4 . . . 3 2 1 2014 ) is a multiple of 5 2015^{(2014^{...^{3^{2^1}}}-2014)}\text{ is a multiple of } 5

The last line is obvious.

Now, just use any big number calculator gives you the 2016th digit =4.

Abdelhamid Saadi
Dec 19, 2016

We can see that n > = 403 we have 201 6 n 201 6 n + 5 2015 m o d 1 0 2016 \forall n >= 403 \quad \text{we have} \quad 2016^n \equiv 2016^{n + 5^{2015}} \mod 10^{2016}

Then 201 6 201 5 201 4 201 3 . . . . 3 2 1 201 6 5 2015 m o d 1 0 2016 \large 2016^{2015^{2014^{2013^{..^{..^{3^{2^{1}}}}}}}} \equiv 2016^{5^{2015}} \mod 10^{2016}

Big calculator the result is 4

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