anonymous
  • anonymous
How do I check 12321 ≡ 111 (mod 3) is true without applying the definition?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
12321 = 111^2 btw
mathmath333
  • mathmath333
as far as i see both are divisible \(3\)
anonymous
  • anonymous
yeah, but that's more like by inspection. I need to use properties of congruences

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ganeshie8
  • ganeshie8
12321 and 111 are divisible by 3 is sufficient
ganeshie8
  • ganeshie8
\(12321\equiv 0 \pmod {3}\) and \(111\equiv 0\pmod{3}\) \(\implies 12321 \equiv 111\pmod{3}\)
anonymous
  • anonymous
by transitive property?
ganeshie8
  • ganeshie8
Yep \(a\equiv c\pmod{n}\) and \(c\equiv b\pmod{n}\) \(\implies a\equiv b\pmod{n}\)
anonymous
  • anonymous
ah yes. I guess this was too easy. Let me pick a different problem. 12345678987654321 ≡ 0 (mod 12345678) ^^
anonymous
  • anonymous
It's according to wolfram alpha, But 12345678987654321 ≡ 0 (mod 12345679) is true though
anonymous
  • anonymous
it's *false* according...
anonymous
  • anonymous
12345678987654321 / 12345679 = 999,999,999
anonymous
  • anonymous
This problem seems challenging but it's in the section where it covers algebraic properties of congruences (addition,subtract and such...). I'm just assuming the exercise is meant for us to apply those properties but it doesn't seem like it.
anonymous
  • anonymous
Maybe the exercises are meant to check the readers' understanding of the definition. Who knows.

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