Last digit of large powers
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
asked Mar 11 at 14:06
AC97AC97
562
$endgroup$
add a comment |
$begingroup$
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
asked Mar 11 at 14:06
AC97AC97
562
$endgroup$
$begingroup$
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
Mar 11 at 15:34
add a comment |
$begingroup$
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
asked Mar 11 at 14:06
AC97AC97
562
$endgroup$
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
asked Mar 11 at 14:06
AC97AC97
562
asked Mar 11 at 14:06
AC97AC97
562
asked Mar 11 at 14:06
AC97AC97
562
asked Mar 11 at 14:06
AC97AC97
562
asked Mar 11 at 14:06
AC97AC97
562
562
$begingroup$
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
Mar 11 at 15:34
add a comment |
$begingroup$
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
Mar 11 at 15:34
$begingroup$
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
Mar 11 at 15:34
$begingroup$
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
Mar 11 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $n ge 2$, $b_n equiv 7^b_n-1 equiv 7^3 equiv 43 (bmod 100)$.
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
$endgroup$
$begingroup$
@ 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
Mar 11 at 15:29
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
add a comment |
$begingroup$
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 $a_2009?$
answered Mar 11 at 15:12
user376343user376343
3,9584829
$endgroup$
$begingroup$
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
Mar 11 at 15:26
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $n ge 2$, $b_n equiv 7^b_n-1 equiv 7^3 equiv 43 (bmod 100)$.
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
$endgroup$
$begingroup$
@ 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
Mar 11 at 15:29
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
add a comment |
$begingroup$
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 $n ge 2$, $b_n equiv 7^b_n-1 equiv 7^3 equiv 43 (bmod 100)$.
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
$endgroup$
$begingroup$
@ 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
Mar 11 at 15:29
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
add a comment |
$begingroup$
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 $n ge 2$, $b_n equiv 7^b_n-1 equiv 7^3 equiv 43 (bmod 100)$.
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
$endgroup$
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 $n ge 2$, $b_n equiv 7^b_n-1 equiv 7^3 equiv 43 (bmod 100)$.
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
answered Mar 11 at 15:26
Robert IsraelRobert Israel
331k23220475
331k23220475
$begingroup$
@ 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
Mar 11 at 15:29
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
add a comment |
$begingroup$
@ 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
Mar 11 at 15:29
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
$begingroup$
@ 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
Mar 11 at 15:29
$begingroup$
@ 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
Mar 11 at 15:29
1
1
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
$begingroup$
The last two digits of $a_2009$ are $07$, and the last two digits of $b_2009$ are $43$.
$endgroup$
– Robert Israel
Mar 11 at 15:34
1
1
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
You're missing a lot of steps in your proofs.
$endgroup$
– Acccumulation
Mar 11 at 17:38
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
$begingroup$
@RobertIsrael How would you go about finding the last digit of $a_2009$ and $b_2009$? I was more confused with that part funnily enough.
$endgroup$
– AC97
Mar 11 at 21:17
add a comment |
$begingroup$
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 $a_2009?$
answered Mar 11 at 15:12
user376343user376343
3,9584829
$endgroup$
$begingroup$
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
Mar 11 at 15:26
add a comment |
$begingroup$
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 $a_2009?$
answered Mar 11 at 15:12
user376343user376343
3,9584829
$endgroup$
$begingroup$
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
Mar 11 at 15:26
add a comment |
$begingroup$
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 $a_2009?$
answered Mar 11 at 15:12
user376343user376343
3,9584829
$endgroup$
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 $a_2009?$
answered Mar 11 at 15:12
user376343user376343
3,9584829
answered Mar 11 at 15:12
user376343user376343
3,9584829
answered Mar 11 at 15:12
user376343user376343
3,9584829
answered Mar 11 at 15:12
user376343user376343
3,9584829
3,9584829
$begingroup$
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
Mar 11 at 15:26
add a comment |
$begingroup$
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
Mar 11 at 15:26
$begingroup$
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
Mar 11 at 15:26
$begingroup$
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
Mar 11 at 15:26
add a comment |
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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
Mar 11 at 15:34