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What is the remainder when $7^{7^{7^{7.........Infinity }}}$ is divided by $5$ ?

My try :

$7^7$ when divided by $5$ gives the remainder $3$,and

similarly, $7^{7^7}$ when divided by $5$ again gives the remainder $3$,and

so, i know that upto infinity it will give remainder $3$.

But, Above approach does not shows any real Maths. How to approach for such questions ?

Edit :

Can I stop the power tower to something like $7^{4k + R}$ ?

Jyrki Lahtonen
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Jon Garrick
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    The remainder is defined when you divide a natural number $n$ to a natural number $m$. However, $7^{7^{7^{7^{...}}}}$ is not a natural number. – Levent Dec 31 '16 at 18:16
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    As @Levent says, you can continue the "power tower" as far up as you want, but it has to stop somewhere. – Arthur Dec 31 '16 at 18:18
  • If I power something ending with "1" infinite times I am pretty sure it will still have "1" on the end – Markoff Chainz Dec 31 '16 at 18:21
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    "Above approach does not shows any real Maths. How to approach for such questions ?" Note that $(x_n)$ defined by $x_0=7$ and $x_{n+1}=7^{x_n}$ is such that $x_n=3\bmod{5}$ for every $n\geqslant1$. Now you have a precise mathematical statement, which you will know how to prove, and which involves no undefined operation and/or quantity. – Did Dec 31 '16 at 18:22
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    Try to power $11$ infinite times. Is it still ending with $1$? – Levent Dec 31 '16 at 18:23
  • Reminds me of the most voted answer to this. – RandomStudent Dec 31 '16 at 18:23
  • Levent, it is infinity with 1 on its end – Markoff Chainz Dec 31 '16 at 18:25
  • In the case of powering $11$, the sequence does not have a limit, therefore it does not make any sense to say it is infinity with $1$ on its end. $10^{-n}$ has $n-1$ zeroes after $0.$ and $1$ in the end. If you take the limit, however, the $1$ disappears. – Levent Dec 31 '16 at 18:27
  • You should remove the $7^{...Infinity}$. As pointed out three times above this is not a sensible expression. As you sure you don't mean $7^{7^{\ldots 7}}$ with $n$ $7$'s instead (which is the defined as a simple recursion in Did's comment)? If you don't fix the question then it's very likely going to be closed. – Winther Dec 31 '16 at 18:51

5 Answers5

3

First off, notice that "$7^{7^{\cdots\infty}}$" is not a number - what does it mean to evaluate an infinite power tower?

What I think you mean is most easily expressed using the operation of tetration. Let $^n7$ denote a length-$n$ power tower of $7$s (so $^37 = 7^{7^7}$). Then the question seems to be "what is the remainder of $^n7$ when divided by $5$, for any $n$?"

My other concern is with your argument - two examples does not a proof make. I could similarly argue "$3$ is prime, $5$ is prime, therefore every odd number is prime". What you want to do is use induction: show that if $^n7$ has remainder $3$ when divided by $5$, then so does $^{n+1}7$. That's actually pretty easy - an answer posted while I was typing this gave a good explanation. But once you have that argument, and the fact that $^27 = 7^7$ works, you now know that $^n7$ has remainder $3$ for every $n > 1$.

Reese Johnston
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$7^1\equiv2\pmod5$

$7^2\equiv4\pmod5$

$7^3\equiv3\pmod5$

$7^4\equiv1\pmod5$

Thus,

$7^n=7^{4k+(n\mod4)}=(7^4)^k\times7^{n\mod4}\equiv1^k\times7^{n\mod4}=7^{n\mod4}\pmod5$

Via similar reasoning,

$7^n=7^{2k+(n\mod2)}=(7^2)^k\times7^{n\mod2}\equiv1^k\times7^{n\mod2}=7^{n\mod2}\pmod4$

Finally,

$$7^{7^{7^n}}\equiv7^{(7^{(7^n\mod2)}\mod4)}=7^3\equiv3\pmod5$$

Simply Beautiful Art
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let $a_1=7$ and let $a_{n+1}=7^{a_n}$. It is clear that $a_n\equiv 3 \bmod 4$ because $7$ is congruent to $3\bmod 4$, and we are raising this number to an odd exponent.

Therefore, by fermats theorem we have $7^{a_n}\equiv 7^3\equiv 3 \bmod 5$.

So $a_n\equiv 3\bmod 5$ for $n>1$.

Asinomás
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$7\equiv 2 \bmod 5$

$2^x \equiv 2^{x \bmod \phi(5)} \bmod 5$

$2^x \equiv 2^{x \bmod 4} \bmod 5$

$7 \equiv 3\bmod 4$

$3^x \equiv -1^{x} \equiv -1^{x\bmod 2} \bmod 4$

$7^{\text{anything}} \equiv 1 \bmod 2$

$7^{7^{\text{anything}}} \equiv 3^{7^{\text{anything}}} \equiv 3 \bmod 4$

$7^{7^{7^{\text{anything}}}} \equiv 2^{3^{7^{\text{anything}}}} \equiv 2^{3} \equiv 3 \bmod 5$

Joffan
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${\rm mod}\,\ \color{#c00}4\!:\,\ \color{#0a0}{7^{\large K}}\!\equiv (-1)^{\large K}\!\equiv\, \color{#c00}{\pm 1},\ $ i.e. $\,+1 $ for even $K;\,$ $\,-1\,$ for odd $K.\,$ Therefore

${\rm mod}\,\ 5\!:\ 7^{\Large\color{#0a0}{7^{\LARGE K}}}\!\!\!\equiv 2^{\large\color{#c00}{\pm1+4N}}\!\equiv 2^{\large \pm1} {\underbrace{(2^{\large 4})}_{\large \equiv\: 1\ }}^{\!\large N}\!\!\equiv 2^{\large \pm1}\!\equiv \pm 2,\ $ i.e. $\, 2\,$ for even $K;\,$ $-2\,$ for odd $K$ (OP)

Bill Dubuque
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