Question

2n+6m≅ 2 (mod 10)
n+m≅ -2 (mod 10)
how to solve this??? n and m

Answer 1

Not sure but I'd start with considering the two different cases

(A) n+3m≅1(mod 10)
and
(B) n+3m≅6(mod 10)

for (A), subtracting the second equation from the first:

2m≅3(mod 10)

which gives no solution.

For (B):

2m≅8(mod 10), giving m≅4 or m≅9 and then you can use the second equation to solve for n (mod 10).

So it's pretty much like solving an system of equations, but you have to be careful with division. In particular dividing with 2 will you either 2 or no solutions.

Answer 2

it's the way i thought it supposed to be ~ well i'll keep trying that~ thx

Answer 3

i should work this way

Answer 4

BTW, can you solve 2k≅4(mod 10) ?

Answer 5

k=2 k=7 works

Answer 6

Yes. So does any

k ≅2(mod 10)
k ≅7(mod 10)

i.e 12,17,22,27, etc. The solution is a congruence itself. For a system it would be a pair of congruences, like

\[a_1≅ 1 (mod 10), b_1≅ 1 (mod 10)\]
\[a_2≅ 5 (mod 10), b_2≅ 7 (mod 10)\]

etc.

Answer 7

You might already know all this though :)

Answer 8

question solving n and m is the one im still working on~
how did u get (A) n+3m≅1(mod 10)
and (B) n+3m≅6(mod 10) ?

Answer 9

I got them from 2n+6m≅2(mod 10)

I "divided" both sides by 2.

Answer 10

well, what about (B) the second u got~ i cant find the relation btw those two.
ive never done that with a negative sign.

Answer 11

B comes from the first equation too. That might not have been clear.

If n+3m≅6(mod 10) then
2n+6m≅12(mod 10)
which is the same as

2n+6m≅2(mod 10)

So when you look at an equation like 2n+6m≅2(mod 10), you ask, what does this mean for n+3m?

It means either that

n+3m≅1(mod 10)
or
n+3m≅6(mod 10).

And then consider the two cases separately.

Answer 12

wow~ that's awesome!~ thank you-!