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So the other day i asked this question: Repetitive 1-9 pow last digit

Which, well the title speaks for itself, and i got a great answer. However, now i'm onto even deeper problems.

So lets take this one:

$12^{30^{21}}$

12 mod 10 is 2 which gives me period 4. That gives me 30 on mod 4

$30^{1}$mod 4 = 2, $30^{2}$mod 4 = 4, $30^{3}$ = 0, $30^{4}$ = 4, $30^{5}$ = 4, $30^{6}$ = 0 and the rest is also 0...

I then make the assumption that 30 mod 4 gives me period 6

21 mod 6 gives me 3.

Then we got 2^2^3 mod 10 which is 4, when in fact it should be 6. What am i doing wrong here?

Also the power of 30 is quite simple, but what happens when higher numbers occur?

$937640^{767456^{98124}}$ or $937640^{767456^{98124}}$

Vollan
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1 Answers1

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I have no idea where "period 6" comes from. Note that $30\equiv_42$, so any power of $30$ with higher exponent than $1$ is just going to be $0\equiv_44$. Thus $30^{21}\equiv_44$, which gives $$12^{30^{21}}\equiv_{10} 2^{30^{21}}\equiv_{10}2^4\equiv_{10}6$$ As for larger numbers, just reduce all bases modulo whatever you're working with, and everything works out. In your "larger numbers" examples (which are identical, by the way), they end in a $0$, so of course any positive integer power of that is also going to end in a $0$. But if we were to change it a bit, for the sake of example, we would do $$ 937643^{767455^{98124}}\equiv_{10} 3^{767455^{98124}} $$ Now, the powers of $3$ modulo $10$ are $3, 9, 7, 1, 3, 9, 7, 1, \ldots$, which is a period of $4$. Thus we are interested in the exponent modulo $4$. We have $767455\equiv_4 -1$, which gives us $$ 767455^{98124} \equiv_4 (-1)^{98124} $$ And the result of raising $-1$ to any power is just dependent on whether the exponent is even or odd. In this case it's even, so we get $$ (-1)^{98124}\equiv_41 $$ which means that we have $$ 3^{767455^{98124}}\equiv_{10} 3^{(-1)^{98124}}\equiv_{10}3^1 $$

Arthur
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  • how is $30^{21}\equiv_44$ also how is $0\equiv_44$ ? – Vollan Dec 19 '19 at 16:32
  • And also 767455 mod 4 is 3? Are we for some reason taking the mod 1 step further than it's supposed to? – Vollan Dec 19 '19 at 16:35
  • We are repeatedly applying Euler and or Carmichael's totient functions. or observation. –  Dec 19 '19 at 17:18
  • @Vollan If you don't know why $4\equiv 0\pmod 4$, then you really ought to go back to the very basics of modular arithmetic and what $\equiv$ actually means. Then it will also be clear to you why $30^{21} = (2\cdot 15)^{21} = 2^{15}\cdot 15^{19}$ is congruent to $0$. I chose to use $4$ rather than $0$ because using $0$ as exponent can be iffy in modular arithmetic, and I didn't want to deal with it. And finding $767455\equiv_4 3$ is almost trivial. Try to show that $767456$ is divisible by $4$, and take it from there. – Arthur Dec 19 '19 at 17:39
  • @Arthur oh, i thought $0\equiv_44$ ment 0 mod 4, which is 0. but yeah, i agree that 4 mod 4 is 0 – Vollan Dec 19 '19 at 18:25
  • @Vollan You're thinking about mod as an operation. I am thinking about it as an equivalence. I think that in number theory, the latter is better. – Arthur Dec 19 '19 at 18:48
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    @Vollan See e.g. here for discussion on mod as a congruence equivalence relation vs. a normal-form operator. Congruences are more more flexible, e.g. we have the freedom to use $,10\equiv -1\pmod{11}$ when casting out elevens or $\rm\color{#c00}{twiddling}$ fractions to get exact quotients – Bill Dubuque Dec 19 '19 at 18:57