Subcategories

• Module 5 Day 12 Challenge Part 1

  Solving Three Congruences

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• Module 5 Day 12 Challenge Part 2

  Reduction to Two Congruences

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  P

  @potato2017 I used the exact same method here, and it was pretty quick.

• Module 5 Day 12 Challenge Part 3

  Table Visualization of Chinese Remainder Theorem

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• Module 5 Day 12 Challenge Part 4

  Table Visualization of Chinese Remainder Theorem

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• Module 5 Day 12 Challenge Part 5

  Table Visualization of Chinese Remainder Theorem

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  E

  Actually I think there is a new way to do this; just let == be congruence, and p be the number of pennies. We know that

  p == 3(mod 4)
  p == 16(mod 25)

  Subtracting 12 on 3(mod 4) gives us p == -9(mod 4), and subtracting 25 on 16(mod 25) gives us p == -9(mod 25), so that means that p == -9(mod 100) which is the same as p == 91(mod 100).

  😁 😁 😁 😁

• Module 5 Day 12 Your Turn Part 1

  Euler's Totient For a Number With Three Prime Factors

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• Module 5 Day 12 Your Turn Part 2

  Euler's Totient For a Number With Three Prime Factors

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• Module 5 Day 12 Your Turn Part 3

  Shortcut Using Chinese Remainder Theorem

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• Module 5 Day 12 Your Turn Part 4

  Another Derivation of Euler's Totient Function

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• Module 5 Day 12 Your Turn Part 5

  Another Derivation of Euler's Totient Function

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