A problem of Modular Arithmetic

[rakeen]

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 ?

[AntiviruShahriar]

$3a+1=n=4b+2 theke n1=10,22..........[10+g12]$ 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????????????????

[Avik Roy]

The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?

[Moon]

$3a+1=n=4b+2 theke n1=10,22..........[10+g12]$ 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????????????????

If you don't write equations correctly, they can become unreadable. I recommend to check your post once after you submit.

[Muhiminul Maruf]

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???????????

[Labib]

I think there's no definite residue!!!
it's 0,1,2,3,4,5,6.

[AntiviruShahriar]

The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?

amar $60n-2$ ashche ken????

[rakeen]

@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 Roy]

The numbers are of the form $60n+1$
There's no single residue when dividing by 7, can there be?

amar $60n-2$ ashche ken????

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 solutions

[AntiviruShahriar]

@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.

asholee ami kono theory use kori nai....shorto dhoree dhoree agaisi r ans peye shetake dharay porinoto korsiii