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anonymous

  • one year ago

How do I check 12321 ≡ 111 (mod 3) is true without applying the definition?

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  1. anonymous
    • one year ago
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    12321 = 111^2 btw

  2. mathmath333
    • one year ago
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    as far as i see both are divisible \(3\)

  3. anonymous
    • one year ago
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    yeah, but that's more like by inspection. I need to use properties of congruences

  4. ganeshie8
    • one year ago
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    12321 and 111 are divisible by 3 is sufficient

  5. ganeshie8
    • one year ago
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    \(12321\equiv 0 \pmod {3}\) and \(111\equiv 0\pmod{3}\) \(\implies 12321 \equiv 111\pmod{3}\)

  6. anonymous
    • one year ago
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    by transitive property?

  7. ganeshie8
    • one year ago
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    Yep \(a\equiv c\pmod{n}\) and \(c\equiv b\pmod{n}\) \(\implies a\equiv b\pmod{n}\)

  8. anonymous
    • one year ago
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    ah yes. I guess this was too easy. Let me pick a different problem. 12345678987654321 ≡ 0 (mod 12345678) ^^

  9. anonymous
    • one year ago
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    It's according to wolfram alpha, But 12345678987654321 ≡ 0 (mod 12345679) is true though

  10. anonymous
    • one year ago
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    it's *false* according...

  11. anonymous
    • one year ago
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    12345678987654321 / 12345679 = 999,999,999

  12. anonymous
    • one year ago
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    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.

  13. anonymous
    • one year ago
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    Maybe the exercises are meant to check the readers' understanding of the definition. Who knows.

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