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For an exercise I had to find the second last number of $7^{(54^{1000} +2)}$ in the number system with base 9. I computed this number modulo $9^2 = 81$. This is what I found:

Since $\text{gcd}(7, 81) = 1$ and $\phi(81) = 54$ (where $\phi$ is the Euler totient function), I found that $$54^{1000} + 2 \equiv 2 \mod 54.$$

Therefore we have that $7^{(54^{1000} + 2)} \equiv 7^2 \equiv 49 \mod 81$.

Now this solution made me suspicious (because of the 9 appearing in my solution). Do I have to convert $49 = 5 \cdot 9 + 4 \cdot 1$ to find the second last digit in the number system with base 9? Or is $49$ the solution?

  • 0
    The last digit is indeed $5$ – 2017-02-18
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    so I had to convert my solution into base 9? Thank you very much! – 2017-02-18

1 Answers 1

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The computations are correct and, yes, you eventually have to convert into base $9$ representation, so $\mathtt{54}$ are the last two digits in base $9$.