Finding the last digit of $103^{103^{103^{103^{103}}}}$

Question

I need to find the last digit of $103^{103^{103^{103^{103}}}}$ so the value in $\mod10$. I know

\begin{align} 103^{103^{103^{103^{103}}}}&=(100+3)^{103^{103^{103^{103}}}}\\ &=100\cdot(stuff)+3^{103^{103^{103^{103}}}}\\ &=3^{103^{103^{103^{103}}}} \mod10 \end{align}

And then I am stuck. I have observed that the last digit of a power of 3 is cyclic: $$3 \to 9 \to 7 \to 1 \to 3 \to 9 \to 7 \to\cdots$$

This means it depends on $103^{103^{103^{103}}}\mod4$.

Is this the right method? do I just need to keep doing this or is there a simpler "trick"?

Isn't the pattern for 103 modulo 4 just 1,3,1,3,1,3 so you need modulo 2, but that's just 1,1,1,1,1,1 so you're done? — Kitter Catter, Dec 03 '19 at 20:18
indeed Wikipedia says the usual rule is to work from the top down — J. W. Tanner, Dec 03 '19 at 21:51
Yes, that's the best general method - see the link on modular order reduction in my answer. — Bill Dubuque, Dec 05 '19 at 05:58
Observing the period four sequence of remainders is a good start. See for example this for ideas on how to continue. Many techniques for problems like this are covered in this mother thread and the close to 200 threads linked to it. — Jyrki Lahtonen, Dec 05 '19 at 06:06

David Pement · Accepted Answer · 2019-12-03T21:18:35.720

3

Hint: $103=-1\pmod4$, and $-1$ to any odd power is $-1$.

Edit: Full answer: once you arrive at $3^{103^{103^{103^{103}}}}$, you can take the exponent mod 4, like you suggested. Drawing from my hint, we have $103^{103^\text{positive int}}=(-1)^{\text{odd number}}=-1=3\pmod 4$, since the exponent $103^\text{positive int}$ is an odd number. So your answer is $3^3=7\pmod{10}$. The last digit is $7$.

edited Dec 03 '19 at 21:18

answered Dec 03 '19 at 20:22

David Pement

836

Edited for clarification. – David Pement Dec 03 '19 at 21:12
odd to any power is odd @aradarbel10$ in fact parity doesn't chsnge under exponentiation. – Dec 03 '19 at 21:17

J. W. Tanner · Answer 2 · 2019-12-03T21:36:19.160

2

You're on the right track.

Your observed pattern can be explained by $3^4=81\equiv1\pmod{10}$.

$103^{103}\equiv3^3\equiv3\pmod{4}$, so $103^{103^{103}} \equiv103^3\equiv3^3\equiv7\pmod{10}$.

$103^{103^{103}}\equiv3^3\equiv3\pmod4,$ so $103^{103^{103^{103}}}\equiv103^3\equiv3^3\equiv7\pmod{10}.$

Can you observe another pattern now?

edited Dec 03 '19 at 21:36

answered Dec 03 '19 at 20:22

J. W. Tanner

60,406

Can't use that $103\equiv 1$ mod $(2)$? – Fareed Abi Farraj Dec 03 '19 at 20:26
why is 103^103≡3^3? am I allowed to change the exponent according to the (mod 10)? – aradarbel10 Dec 03 '19 at 20:27
@aradarbel10: $103^{103}\equiv 3^{103}=(3^4)^{25}3^3\equiv3^3$; when working modulo $10$, reduce the exponent modulo $4$ – J. W. Tanner Dec 03 '19 at 20:28
@J.W.Tanner and, sorry for asking, is $103^7$ because of the $103^{103^{103}}=103^{100\cdot(\cdots)+7}$ ? – aradarbel10 Dec 03 '19 at 20:50
Don't be sorry, @aradarbel10; I think I made a mistake there and have edited to correct now – J. W. Tanner Dec 03 '19 at 21:46

Bill Dubuque · Answer 3 · 2020-05-19T20:56:04.160

$\color{#0a0}{k\ \rm odd},\ (a,10)\!=\!1\overset{\rm Euler}\Rightarrow\bmod 10\!:\ a^{\large\color{#c00} 4}\equiv 1\,\overset{\rm\color{#90f}{MOR}}\Rightarrow\, a^{\large (4n-1)^{\Large k}}\!\!\equiv a^{\large (\overbrace{4n-1)^{\Large k}\bmod\color{#c00} 4}^{\!\!\!\!\!\Large \equiv\ \,\color{#0a0}{(-1)^{k}\ \equiv}\,\ \LARGE\color{#0a0}3 }}\!\equiv a^{\large \color{#0a0}3}$

by Euler's $\phi$ Theorem combined with $\,\rm\color{#90f}{MOR}= \,$ Modular Order Reduction.

The OP is the special case $\, a = 103 = 4n\!-\!1,\,\ k = 103^{103^{103}}\! $ (or any odd integer)

Finding the last digit of $103^{103^{103^{103^{103}}}}$

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