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Pattern in power towers of 2 involving last digits

Fermat: Last two digits of $7^355$$a^100-1$ is divisible by $1000$.Is there a quicker proof to show that $2^10^k equiv 7 pmod9$ for all positive integers $k$?Attempt on fractional tetrationChecking my work for $21^100-12^100$ modulo $11$Finding the remainder of a factorial using modular arithmeticFormula for consecutive residue of primitive modulo n.Proof Involving Modular Arithmetic and Fermat's TheoremHow to tell if system of congruences where each base is a power of prime $p$ has a solutionFind $21^1234pmod100equiv ?$

1

$begingroup$

We have
beginalign
2^2^2 &mod 10 = 6 \
2^2^2^2 &mod 100 = 36 \
2^2^2^2^2 &mod 1000 = 736 \
2^2^2^2^2^2 &mod 10000 = 8736 \
2^2^2^2^2^2^2 &mod 100000 = 48736
endalign

I think you get the point. Basically, it seems like $^n2 equiv ^n+12 mod 10^n-2$ for $n geq 3$, where $^n 2$ represents tetration. How could one go about proving this?

modular-arithmetic tetration power-towers

share|cite|improve this question

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

$endgroup$

add a comment | 

1

$begingroup$

We have
beginalign
2^2^2 &mod 10 = 6 \
2^2^2^2 &mod 100 = 36 \
2^2^2^2^2 &mod 1000 = 736 \
2^2^2^2^2^2 &mod 10000 = 8736 \
2^2^2^2^2^2^2 &mod 100000 = 48736
endalign

I think you get the point. Basically, it seems like $^n2 equiv ^n+12 mod 10^n-2$ for $n geq 3$, where $^n 2$ represents tetration. How could one go about proving this?

modular-arithmetic tetration power-towers

share|cite|improve this question

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

$endgroup$

add a comment | 

$begingroup$

We have
beginalign
2^2^2 &mod 10 = 6 \
2^2^2^2 &mod 100 = 36 \
2^2^2^2^2 &mod 1000 = 736 \
2^2^2^2^2^2 &mod 10000 = 8736 \
2^2^2^2^2^2^2 &mod 100000 = 48736
endalign

I think you get the point. Basically, it seems like $^n2 equiv ^n+12 mod 10^n-2$ for $n geq 3$, where $^n 2$ represents tetration. How could one go about proving this?

modular-arithmetic tetration power-towers

share|cite|improve this question

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

$endgroup$

share|cite|improve this question

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

share|cite|improve this question

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

edited Mar 16 at 23:20

Peter Foreman

4,1771216

edited Mar 16 at 23:20

Peter Foreman

4,1771216

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

asked Mar 16 at 23:13

Davidmath7Davidmath7

1665

1 Answer
1

active

oldest

votes

2

$begingroup$

Let $x_0=4$ and $x_n+1=2^x_n$.
The claim is $x_n+1equiv x_npmod10^n$.
By induction on $n$, we assume $x_n+1equiv x_npmod10^n$ and we prove $x_n+2equiv x_n+1pmod10^n+1$.

We have
$$x_n+2-x_n+1=2^x_n+1-2^x_n=2^x_n(2^x_n+1-x_n-1)$$
Since $x_ngeq n+1$ we get $x_n+2equiv x_n+1pmod2^n+1$.
On the other hand for $ngeq 2$ we have $x_n+1equiv x_npmod4$ and $x_n+1equiv x_npmod5^n$ by assumption, so that $x_n+1-x_nequiv 0pmod4cdot 5^n$.
Since $varphi(5^n+1)=4cdot 5^n$, this gives $2^x_n+1-x_nequiv 1pmod5^n+1$, thus giving $x_n+2equiv x_n+1pmod5^n+1$.
This, together with $x_n+2equiv x_n+1pmod2^n+1$ gives $x_n+2equiv x_n+1pmod10^n+1$ concluding the proof.

share|cite|improve this answer

edited Mar 16 at 23:37

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

$endgroup$

add a comment | 

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2

$begingroup$

Let $x_0=4$ and $x_n+1=2^x_n$.
The claim is $x_n+1equiv x_npmod10^n$.
By induction on $n$, we assume $x_n+1equiv x_npmod10^n$ and we prove $x_n+2equiv x_n+1pmod10^n+1$.

We have
$$x_n+2-x_n+1=2^x_n+1-2^x_n=2^x_n(2^x_n+1-x_n-1)$$
Since $x_ngeq n+1$ we get $x_n+2equiv x_n+1pmod2^n+1$.
On the other hand for $ngeq 2$ we have $x_n+1equiv x_npmod4$ and $x_n+1equiv x_npmod5^n$ by assumption, so that $x_n+1-x_nequiv 0pmod4cdot 5^n$.
Since $varphi(5^n+1)=4cdot 5^n$, this gives $2^x_n+1-x_nequiv 1pmod5^n+1$, thus giving $x_n+2equiv x_n+1pmod5^n+1$.
This, together with $x_n+2equiv x_n+1pmod2^n+1$ gives $x_n+2equiv x_n+1pmod10^n+1$ concluding the proof.

share|cite|improve this answer

edited Mar 16 at 23:37

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

$endgroup$

add a comment | 

2

$begingroup$

Let $x_0=4$ and $x_n+1=2^x_n$.
The claim is $x_n+1equiv x_npmod10^n$.
By induction on $n$, we assume $x_n+1equiv x_npmod10^n$ and we prove $x_n+2equiv x_n+1pmod10^n+1$.

We have
$$x_n+2-x_n+1=2^x_n+1-2^x_n=2^x_n(2^x_n+1-x_n-1)$$
Since $x_ngeq n+1$ we get $x_n+2equiv x_n+1pmod2^n+1$.
On the other hand for $ngeq 2$ we have $x_n+1equiv x_npmod4$ and $x_n+1equiv x_npmod5^n$ by assumption, so that $x_n+1-x_nequiv 0pmod4cdot 5^n$.
Since $varphi(5^n+1)=4cdot 5^n$, this gives $2^x_n+1-x_nequiv 1pmod5^n+1$, thus giving $x_n+2equiv x_n+1pmod5^n+1$.
This, together with $x_n+2equiv x_n+1pmod2^n+1$ gives $x_n+2equiv x_n+1pmod10^n+1$ concluding the proof.

share|cite|improve this answer

edited Mar 16 at 23:37

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

$endgroup$

add a comment | 

$begingroup$

Let $x_0=4$ and $x_n+1=2^x_n$.
The claim is $x_n+1equiv x_npmod10^n$.
By induction on $n$, we assume $x_n+1equiv x_npmod10^n$ and we prove $x_n+2equiv x_n+1pmod10^n+1$.

We have
$$x_n+2-x_n+1=2^x_n+1-2^x_n=2^x_n(2^x_n+1-x_n-1)$$
Since $x_ngeq n+1$ we get $x_n+2equiv x_n+1pmod2^n+1$.
On the other hand for $ngeq 2$ we have $x_n+1equiv x_npmod4$ and $x_n+1equiv x_npmod5^n$ by assumption, so that $x_n+1-x_nequiv 0pmod4cdot 5^n$.
Since $varphi(5^n+1)=4cdot 5^n$, this gives $2^x_n+1-x_nequiv 1pmod5^n+1$, thus giving $x_n+2equiv x_n+1pmod5^n+1$.
This, together with $x_n+2equiv x_n+1pmod2^n+1$ gives $x_n+2equiv x_n+1pmod10^n+1$ concluding the proof.

share|cite|improve this answer

edited Mar 16 at 23:37

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

$endgroup$

share|cite|improve this answer

edited Mar 16 at 23:37

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426

answered Mar 16 at 23:32

Fabio LucchiniFabio Lucchini

9,46111426