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.







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  • $begingroup$
    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











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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.










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