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I am trying to find the unit digit of ${{43}^{43}}^{43}$. Normally I would try to break up the exponent to make it easier though 43 is prime so I can't do that. So instead I first considered $43^{43}$ and tried finding a pattern for the unit digits by looking at the first few values:

$43^0=1$

$43^1=43$

$43^2=1849$

$43^3=79507$

$43^4=3418801$

$43^5=147008443$

...

I then noticed that they repeated every 4 values with a unit digit pattern of {1, 3, 9, 7}.

This is where I am stuck, for a few reasons:

  • I know I need to prove that this unit digit pattern continues but I'm not quite sure how to do this.
  • I know the unit digit is 1, 3, 7 or 9, but how do I know which one it is?
  • After finding the unit digit of $43^{43}$, how do I then use this to find the unit digit of ${{43}^{43}}^{43}$? Lets say $x$ is the unit digit of $43^{43}$ so would I then need to find the unit digit of $43^x$?

Can someone help me with where I am confused, let me know if the proccess I wrote is incorrect, or explain an easier approach to this problem?

AMM
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    Hint: finding the last digit is the same as simplify the expression modulo 10 – Exodd May 19 '20 at 20:40
  • Note the units repeat on a cycle of length $4$. Since $4|43^{43}+1$, $10|43^{43^{43}}-43^3$, making the units digit $7$. – J.G. May 19 '20 at 20:51
  • Special case of this answer in the dupe. See also the many linked questions here for many further examples. – Bill Dubuque May 19 '20 at 20:51
  • $43\equiv3\bmod10$ and $3^4\equiv1\bmod10$, so $3^{4k+m}\equiv 3^m\bmod10$. And $43^{43}\equiv(-1)^{43}=-1\equiv3\bmod4$, so $43^{43^{43}}\equiv3^3\equiv7\bmod10$ – J. W. Tanner May 19 '20 at 20:55
  • @J.W.Tanner I understand $43$ $≡$ $3$ mod $10$ and that $43^{43}$ $≡$ $(−1)^{43}$ $=$ $−1$ $≡$ $3$ mod $4$ though why can we say ${43^{43}}^{43}$ $≡$ $3^3$ if the first part was for mod $10$ and the other was for mod $4$? – AMM May 22 '20 at 20:50
  • When we're dealing with mod $10$, exponents get reduced mod $\phi(10)=4$; $a^4\equiv1\bmod10$ if $\gcd(a,10)=1$ – J. W. Tanner May 22 '20 at 21:05

1 Answers1

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$43^{43}\equiv (-1)^{43} \equiv 3 \mod \phi(10) =4$, so $43^{43^{43}}$ has last digit $7$ as number $43^3$.

Pavel Kozlov
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