Mixing
last 2 digits of a sequence
Clash Royale CLAN TAG#URR8PPP
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$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
sequences-and-series
asked Aug 2 at 9:25
SuperMage1
53718
add a comment |Â
up vote
4
down vote
favorite
$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
sequences-and-series
asked Aug 2 at 9:25
SuperMage1
53718
The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
2
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
sequences-and-series
asked Aug 2 at 9:25
SuperMage1
53718
$x+frac1x = 3$, what are the last 2 digits of $x^2^2013+frac1x^2^2013$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
sequences-and-series
asked Aug 2 at 9:25
SuperMage1
53718
asked Aug 2 at 9:25
SuperMage1
53718
asked Aug 2 at 9:25
SuperMage1
53718
asked Aug 2 at 9:25
SuperMage1
53718
53718
The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
2
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36
add a comment |Â
The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
2
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36
The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
2
2
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
add a comment |Â
up vote
0
down vote
The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$
That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$
Starting with 3 and working mod $(100)$ the algorithm goes like,
$$3to 7 to 47 to 07 to 47 to 07 to .....$$
Now you can finish the solution.
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
add a comment |Â
up vote
4
down vote
accepted
Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
Definitely you see the pattern : $x^2m + frac 1x^2m = left(x^m + frac 1x^mright)^2 - 2$ for all $m geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22colorblue07$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^2^n + frac 1x^2^n$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
answered Aug 2 at 9:37
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.5k22463
31.5k22463
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
add a comment |Â
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
Well the algebra certainly beats my trial and error explanation
– Mohammad Zuhair Khan
Aug 2 at 9:41
add a comment |Â
up vote
0
down vote
The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$
That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$
Starting with 3 and working mod $(100)$ the algorithm goes like,
$$3to 7 to 47 to 07 to 47 to 07 to .....$$
Now you can finish the solution.
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
up vote
0
down vote
The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$
That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$
Starting with 3 and working mod $(100)$ the algorithm goes like,
$$3to 7 to 47 to 07 to 47 to 07 to .....$$
Now you can finish the solution.
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$
That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$
Starting with 3 and working mod $(100)$ the algorithm goes like,
$$3to 7 to 47 to 07 to 47 to 07 to .....$$
Now you can finish the solution.
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
The trick is $$(a+1/a)^2= a^2 + 1/a^2 +2$$
That is $$ a^2 + 1/a^2=(a+1/a)^2-2$$
Starting with 3 and working mod $(100)$ the algorithm goes like,
$$3to 7 to 47 to 07 to 47 to 07 to .....$$
Now you can finish the solution.
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:44
Mohammad Riazi-Kermani
27.1k41851
27.1k41851
add a comment |Â
add a comment |Â
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The next step is exactly the same as the first step.
– Arthur
Aug 2 at 9:33
2
The last two digits alternate between $07$ and $47$.
– Mohammad Zuhair Khan
Aug 2 at 9:35
math.stackexchange.com/questions/936479/… and math.stackexchange.com/questions/2276746/…
– lab bhattacharjee
Aug 2 at 9:36