- #1

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can we say:

[tex]x\equiv[/tex]2 (mod k)

[tex]x\equiv[/tex]2 (mod m)

hence

[tex]x\equiv[/tex]2 (mod km) by km i mean k multiplied by m.

if not, what is the result? or can it be found?

thank you in advance.

- Thread starter theIBnerd
- Start date

- #1

- 13

- 0

can we say:

[tex]x\equiv[/tex]2 (mod k)

[tex]x\equiv[/tex]2 (mod m)

hence

[tex]x\equiv[/tex]2 (mod km) by km i mean k multiplied by m.

if not, what is the result? or can it be found?

thank you in advance.

- #2

Office_Shredder

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k=4

m=8

x=10

x=2(mod 4) and x=2(mod 8) but x=10(mod 32)

In general, if you have something mod m and something mod k, and want to discuss what happens mod mk, then you need a condition on m and k being coprime, or something similar.

- #3

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i think i found sth:

say (k,m) = 1

x=a (mod k)

x=a (mod m)

x=kt+a and x=my+a

kt=my

t=mb

y=kb

then x=kmb+a

x-a=kmb

x-a=0 (mod km)

x=a (mod km)

it is valid, isnt it? any counterexamples?

- #4

Office_Shredder

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That looks pretty good to me

- #5

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:) then my problem is solved. now i should get back to work.

- #6

HallsofIvy

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Yeah, I hate when that happens!:tongue2:

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