find a number $aabb$ such that is full square [duplicate]
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This question already has an answer here:
Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square
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I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?
divisibility
edited Nov 15 at 21:40
amWhy
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
0
down vote
favorite
This question already has an answer here:
Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square
5 answers
I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?
divisibility
edited Nov 15 at 21:40
amWhy
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square
5 answers
I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?
divisibility
edited Nov 15 at 21:40
amWhy
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
This question already has an answer here:
Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square
5 answers
I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?
This question already has an answer here:
Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square
5 answers
divisibility
divisibility
edited Nov 15 at 21:40
amWhy
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
edited Nov 15 at 21:40
amWhy
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
edited Nov 15 at 21:40
amWhy
191k27223437
edited Nov 15 at 21:40
amWhy
191k27223437
edited Nov 15 at 21:40
amWhy
191k27223437
191k27223437
asked Nov 15 at 18:23
Marko Škorić
65110
asked Nov 15 at 18:23
Marko Škorić
65110
asked Nov 15 at 18:23
Marko Škorić
65110
65110
marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
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up vote
1
down vote
We have that
$$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$
and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.
edited Nov 15 at 21:37
amWhy
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":a mid bwhich renders $amid b$, when formatted in mathjax. Don't usea|b
– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have that
$$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$
and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.
edited Nov 15 at 21:37
amWhy
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":a mid bwhich renders $amid b$, when formatted in mathjax. Don't usea|b
– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
add a comment |
up vote
1
down vote
We have that
$$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$
and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.
edited Nov 15 at 21:37
amWhy
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":a mid bwhich renders $amid b$, when formatted in mathjax. Don't usea|b
– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
add a comment |
up vote
1
down vote
up vote
1
down vote
We have that
$$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$
and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.
edited Nov 15 at 21:37
amWhy
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
We have that
$$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$
and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.
edited Nov 15 at 21:37
amWhy
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
edited Nov 15 at 21:37
amWhy
191k27223437
edited Nov 15 at 21:37
amWhy
191k27223437
edited Nov 15 at 21:37
amWhy
191k27223437
191k27223437
answered Nov 15 at 18:27
gimusi
86.3k74392
answered Nov 15 at 18:27
gimusi
86.3k74392
answered Nov 15 at 18:27
gimusi
86.3k74392
86.3k74392
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":a mid bwhich renders $amid b$, when formatted in mathjax. Don't usea|b
– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
add a comment |
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":a mid bwhich renders $amid b$, when formatted in mathjax. Don't usea|b
– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
It's not immediately clear why you exclude cases of b = 2,3,7,8?
– IanF1
Nov 15 at 19:32
1
1
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
@IanF1 Because $n^2mod 10$ can’t take those values.
– gimusi
Nov 15 at 19:34
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
– IanF1
Nov 15 at 19:35
"a divides b":
a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b– amWhy
Nov 15 at 21:38
"a divides b":
a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b– amWhy
Nov 15 at 21:38
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
@amWhy Thanks for the editing and the suggestion! Bye
– gimusi
Nov 15 at 21:49
add a comment |
