2 0 1 6 2 0 1 5 2 0 1 4 2 0 1 3 . . . . 3 2 1
Find the 2016th rightmost digit of the decimal representation of the number above.
You can use a calculator or computer program.
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We can see that ∀ n > = 4 0 3 we have 2 0 1 6 n ≡ 2 0 1 6 n + 5 2 0 1 5 m o d 1 0 2 0 1 6
Then 2 0 1 6 2 0 1 5 2 0 1 4 2 0 1 3 . . . . 3 2 1 ≡ 2 0 1 6 5 2 0 1 5 m o d 1 0 2 0 1 6
Big calculator the result is 4
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We proved in the solution of Once again 2016 simplified that 2 0 1 6 2 0 1 5 2 0 1 4 ∈ S 2 0 1 5 . Here we complete the solution proving that 2 0 1 6 2 0 1 5 2 0 1 4 . . . 3 2 1 ∈ S 2 0 1 6 .
Note that 2 0 1 6 2 0 1 5 2 0 1 4 ∈ S 2 0 1 5 ⇒ ( 2 0 1 6 2 0 1 5 2 0 1 4 ) 5 = 2 0 1 6 5 ∗ 2 0 1 5 2 0 1 4 ∈ S 2 0 1 6 This means we only need to prove that 2 0 1 5 2 0 1 4 . . . 3 2 1 is a multiple of ( 5 ∗ 2 0 1 5 2 0 1 4 )
⇔
2 0 1 5 ( 2 0 1 4 . . . 3 2 1 − 2 0 1 4 ) is a multiple of 5
The last line is obvious.
Now, just use any big number calculator gives you the 2016th digit =4.