Last digit of large powers












6












$begingroup$


Define the sequences $a_1, a_2,...$ and $b_1, b_2,...$ by $a_1=b_1=7$ and where $a_{n+1} = a_n^7$ and $b_{n+1} = 7^{b_n}$ for $n geq 1$.



1) Find the last digit of $a_{2009}$ and $b_{2009}$



2) What about the last two digits - or more?



Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I was working on this, I found than $a_n$ can be written as $a_n = 7^{(7^{n-1})}$, not sure if this is correct or if it helps.



If anyone can help solve this question or guide me in the right direction, I'd be thankful.






sequences-and-series number-theory elementary-number-theory power-series







share|cite|improve this question







New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
    $endgroup$
    – rabota
    4 hours ago


















6












$begingroup$


Define the sequences $a_1, a_2,...$ and $b_1, b_2,...$ by $a_1=b_1=7$ and where $a_{n+1} = a_n^7$ and $b_{n+1} = 7^{b_n}$ for $n geq 1$.



1) Find the last digit of $a_{2009}$ and $b_{2009}$



2) What about the last two digits - or more?



Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I was working on this, I found than $a_n$ can be written as $a_n = 7^{(7^{n-1})}$, not sure if this is correct or if it helps.



If anyone can help solve this question or guide me in the right direction, I'd be thankful.






sequences-and-series number-theory elementary-number-theory power-series







share|cite|improve this question







New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
    $endgroup$
    – rabota
    4 hours ago
















6












6








6


1



$begingroup$


Define the sequences $a_1, a_2,...$ and $b_1, b_2,...$ by $a_1=b_1=7$ and where $a_{n+1} = a_n^7$ and $b_{n+1} = 7^{b_n}$ for $n geq 1$.



1) Find the last digit of $a_{2009}$ and $b_{2009}$



2) What about the last two digits - or more?



Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I was working on this, I found than $a_n$ can be written as $a_n = 7^{(7^{n-1})}$, not sure if this is correct or if it helps.



If anyone can help solve this question or guide me in the right direction, I'd be thankful.






sequences-and-series number-theory elementary-number-theory power-series







share|cite|improve this question







New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Define the sequences $a_1, a_2,...$ and $b_1, b_2,...$ by $a_1=b_1=7$ and where $a_{n+1} = a_n^7$ and $b_{n+1} = 7^{b_n}$ for $n geq 1$.



1) Find the last digit of $a_{2009}$ and $b_{2009}$



2) What about the last two digits - or more?



Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I was working on this, I found than $a_n$ can be written as $a_n = 7^{(7^{n-1})}$, not sure if this is correct or if it helps.



If anyone can help solve this question or guide me in the right direction, I'd be thankful.






sequences-and-series number-theory elementary-number-theory power-series




sequences-and-series number-theory elementary-number-theory power-series






share|cite|improve this question







New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 6 hours ago









AC97AC97

511




511




New contributor




AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






AC97 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
    $endgroup$
    – rabota
    4 hours ago




















  • $begingroup$
    You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
    $endgroup$
    – rabota
    4 hours ago


















$begingroup$
You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
$endgroup$
– rabota
4 hours ago






$begingroup$
You seem to be asking two questions. You can read about the last digits of $b_n$ here: Last digits of power towers $7$, $7^7$, $7^{7^7}$, $7^{7^{7^7}}$, ... don't change, and generalisation
$endgroup$
– rabota
4 hours ago












2 Answers
2






active

oldest

votes


















2












$begingroup$

The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 equiv 1 mod 100$, so $7^{4i+j} equiv 7^j (bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.



Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} equiv 7^0 = 1 (bmod 4)$, and $a_{2009} equiv 7^{1} = 7 (bmod 100)$.



As for $b_{2009}$, we have $b_1 = 7 equiv 3 (bmod 4)$, and since $3^3 = 27 equiv 3 (bmod 4)$ we get $b_n equiv 3 (bmod 4)$ for all $n ge 1$,
and so for $n ge 2$, $b_n equiv 7^{b_{n-1}} equiv 7^3 equiv 43 (bmod 100)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
    $endgroup$
    – AC97
    4 hours ago










  • $begingroup$
    The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
    $endgroup$
    – Robert Israel
    4 hours ago










  • $begingroup$
    You're missing a lot of steps in your proofs.
    $endgroup$
    – Acccumulation
    2 hours ago



















1












$begingroup$

You're right, the powers of $7$ have the last digit $7-9-3-1.$The last digit of $7^7=a_2$ is $3.$



For the powers of $3$ is the last digit $3-9-7-1.$ So the last digit of $a_3$ is $7.$



Can you finish for $a_{2009}?$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
    $endgroup$
    – AC97
    4 hours ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






AC97 is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143708%2flast-digit-of-large-powers%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 equiv 1 mod 100$, so $7^{4i+j} equiv 7^j (bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.



Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} equiv 7^0 = 1 (bmod 4)$, and $a_{2009} equiv 7^{1} = 7 (bmod 100)$.



As for $b_{2009}$, we have $b_1 = 7 equiv 3 (bmod 4)$, and since $3^3 = 27 equiv 3 (bmod 4)$ we get $b_n equiv 3 (bmod 4)$ for all $n ge 1$,
and so for $n ge 2$, $b_n equiv 7^{b_{n-1}} equiv 7^3 equiv 43 (bmod 100)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
    $endgroup$
    – AC97
    4 hours ago










  • $begingroup$
    The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
    $endgroup$
    – Robert Israel
    4 hours ago










  • $begingroup$
    You're missing a lot of steps in your proofs.
    $endgroup$
    – Acccumulation
    2 hours ago
















2












$begingroup$

The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 equiv 1 mod 100$, so $7^{4i+j} equiv 7^j (bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.



Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} equiv 7^0 = 1 (bmod 4)$, and $a_{2009} equiv 7^{1} = 7 (bmod 100)$.



As for $b_{2009}$, we have $b_1 = 7 equiv 3 (bmod 4)$, and since $3^3 = 27 equiv 3 (bmod 4)$ we get $b_n equiv 3 (bmod 4)$ for all $n ge 1$,
and so for $n ge 2$, $b_n equiv 7^{b_{n-1}} equiv 7^3 equiv 43 (bmod 100)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
    $endgroup$
    – AC97
    4 hours ago










  • $begingroup$
    The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
    $endgroup$
    – Robert Israel
    4 hours ago










  • $begingroup$
    You're missing a lot of steps in your proofs.
    $endgroup$
    – Acccumulation
    2 hours ago














2












2








2





$begingroup$

The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 equiv 1 mod 100$, so $7^{4i+j} equiv 7^j (bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.



Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} equiv 7^0 = 1 (bmod 4)$, and $a_{2009} equiv 7^{1} = 7 (bmod 100)$.



As for $b_{2009}$, we have $b_1 = 7 equiv 3 (bmod 4)$, and since $3^3 = 27 equiv 3 (bmod 4)$ we get $b_n equiv 3 (bmod 4)$ for all $n ge 1$,
and so for $n ge 2$, $b_n equiv 7^{b_{n-1}} equiv 7^3 equiv 43 (bmod 100)$.






share|cite|improve this answer









$endgroup$



The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 equiv 1 mod 100$, so $7^{4i+j} equiv 7^j (bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.



Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} equiv 7^0 = 1 (bmod 4)$, and $a_{2009} equiv 7^{1} = 7 (bmod 100)$.



As for $b_{2009}$, we have $b_1 = 7 equiv 3 (bmod 4)$, and since $3^3 = 27 equiv 3 (bmod 4)$ we get $b_n equiv 3 (bmod 4)$ for all $n ge 1$,
and so for $n ge 2$, $b_n equiv 7^{b_{n-1}} equiv 7^3 equiv 43 (bmod 100)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Robert IsraelRobert Israel

327k23216469




327k23216469












  • $begingroup$
    @ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
    $endgroup$
    – AC97
    4 hours ago










  • $begingroup$
    The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
    $endgroup$
    – Robert Israel
    4 hours ago










  • $begingroup$
    You're missing a lot of steps in your proofs.
    $endgroup$
    – Acccumulation
    2 hours ago


















  • $begingroup$
    @ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
    $endgroup$
    – AC97
    4 hours ago










  • $begingroup$
    The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
    $endgroup$
    – Robert Israel
    4 hours ago










  • $begingroup$
    You're missing a lot of steps in your proofs.
    $endgroup$
    – Acccumulation
    2 hours ago
















$begingroup$
@ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
$endgroup$
– AC97
4 hours ago




$begingroup$
@ Robert Israel So are you saying the last digit of $a_{2009}$ is 7 and the last digit of $b_{2009}$ is 3?
$endgroup$
– AC97
4 hours ago












$begingroup$
The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
$endgroup$
– Robert Israel
4 hours ago




$begingroup$
The last two digits of $a_{2009}$ are $07$, and the last two digits of $b_{2009}$ are $43$.
$endgroup$
– Robert Israel
4 hours ago












$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
2 hours ago




$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
2 hours ago











1












$begingroup$

You're right, the powers of $7$ have the last digit $7-9-3-1.$The last digit of $7^7=a_2$ is $3.$



For the powers of $3$ is the last digit $3-9-7-1.$ So the last digit of $a_3$ is $7.$



Can you finish for $a_{2009}?$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
    $endgroup$
    – AC97
    4 hours ago
















1












$begingroup$

You're right, the powers of $7$ have the last digit $7-9-3-1.$The last digit of $7^7=a_2$ is $3.$



For the powers of $3$ is the last digit $3-9-7-1.$ So the last digit of $a_3$ is $7.$



Can you finish for $a_{2009}?$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
    $endgroup$
    – AC97
    4 hours ago














1












1








1





$begingroup$

You're right, the powers of $7$ have the last digit $7-9-3-1.$The last digit of $7^7=a_2$ is $3.$



For the powers of $3$ is the last digit $3-9-7-1.$ So the last digit of $a_3$ is $7.$



Can you finish for $a_{2009}?$






share|cite|improve this answer









$endgroup$



You're right, the powers of $7$ have the last digit $7-9-3-1.$The last digit of $7^7=a_2$ is $3.$



For the powers of $3$ is the last digit $3-9-7-1.$ So the last digit of $a_3$ is $7.$



Can you finish for $a_{2009}?$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









user376343user376343

3,9334829




3,9334829












  • $begingroup$
    Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
    $endgroup$
    – AC97
    4 hours ago


















  • $begingroup$
    Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
    $endgroup$
    – AC97
    4 hours ago
















$begingroup$
Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
$endgroup$
– AC97
4 hours ago




$begingroup$
Apart from using a calculator, how did you find the last digit of $a_2$ to be 3? Then the last digit of $a_3$ to be 7?
$endgroup$
– AC97
4 hours ago










AC97 is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















AC97 is a new contributor. Be nice, and check out our Code of Conduct.













AC97 is a new contributor. Be nice, and check out our Code of Conduct.












AC97 is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143708%2flast-digit-of-large-powers%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more