A problem of Modular Arithmetic
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Dividing a positive integer by 3 the residue becomes 1
by 4 the residue becomes 2
by 5 the residue becomes 3
by 6 the residue becomes 4
what is the residue if we divide the integer by 7 ?
by 4 the residue becomes 2
by 5 the residue becomes 3
by 6 the residue becomes 4
what is the residue if we divide the integer by 7 ?
r@k€€/|/
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Re: A problem of Modular Arithmetic
$3a+1=n=4b+2 theke n1=10,22..........[10+g12]$ gpurnosongkha
$n1=5c+3 er jonno 58, 118.......{(5x-1)12+10}[n{1} hotee dhara toiri koree]$
${(5x-1)12+10} x=natural number$
eita ki thik ase????????????????
$n1=5c+3 er jonno 58, 118.......{(5x-1)12+10}[n{1} hotee dhara toiri koree]$
${(5x-1)12+10} x=natural number$
eita ki thik ase????????????????
Re: A problem of Modular Arithmetic
The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?
There's no single residue when dividing by 7, can there be?
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
Re: A problem of Modular Arithmetic
If you don't write equations correctly, they can become unreadable. I recommend to check your post once after you submit.AntiviruShahriar wrote:$3a+1=n=4b+2 theke n1=10,22..........[10+g12]$ gpurnosongkha
$n1=5c+3 er jonno 58, 118.......{(5x-1)12+10}[n{1} hotee dhara toiri koree]$
${(5x-1)12+10} x=natural number$
eita ki thik ase????????????????
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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Re: A problem of Modular Arithmetic
suppose the integer is X,then
X _= 1 (mod 3)................1
X _= 2 (mod 4)....................2
X _= 3 (mod 5)........................3
X _= 4 (mod 6)............................4
but how can i make a relation with these four equation???????????
X _= 1 (mod 3)................1
X _= 2 (mod 4)....................2
X _= 3 (mod 5)........................3
X _= 4 (mod 6)............................4
but how can i make a relation with these four equation???????????
Re: A problem of Modular Arithmetic
I think there's no definite residue!!!
it's 0,1,2,3,4,5,6.
it's 0,1,2,3,4,5,6.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
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Re: A problem of Modular Arithmetic
amar $60n-2$ ashche ken????Avik Roy wrote:The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?
Re: A problem of Modular Arithmetic
@Anti tomar every line individually bujhchi. But 1ta line er sathe r ekta line er kono relation khuje pachi na.
@Avik if we take n=1 , then 61 is not divisible by 7. the residue becomes 5. and if I consider Anti's result, for n=1 it is 58. and 58|7 .
@Maruf I also tried that way. But I need same modulo (mod x) for each of the equation to set up a relation. Generally we can guess it might be 5, for 7.
@Avik if we take n=1 , then 61 is not divisible by 7. the residue becomes 5. and if I consider Anti's result, for n=1 it is 58. and 58|7 .
@Maruf I also tried that way. But I need same modulo (mod x) for each of the equation to set up a relation. Generally we can guess it might be 5, for 7.
r@k€€/|/
Re: A problem of Modular Arithmetic
I'm sorry. I just misread the problem (which is quite common with me ). $60n-2$ is the correct one. However, that doesnt change the scenario much, sicne it still offers all available solutionsAntiviruShahriar wrote:amar $60n-2$ ashche ken????Avik Roy wrote:The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
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Re: A problem of Modular Arithmetic
asholee ami kono theory use kori nai....shorto dhoree dhoree agaisi r ans peye shetake dharay porinoto korsiiirakeen wrote:@Anti tomar every line individually bujhchi. But 1ta line er sathe r ekta line er kono relation khuje pachi na.
@Avik if we take n=1 , then 61 is not divisible by 7. the residue becomes 5. and if I consider Anti's result, for n=1 it is 58. and 58|7 .
@Maruf I also tried that way. But I need same modulo (mod x) for each of the equation to set up a relation. Generally we can guess it might be 5, for 7.