Subcategories

• Day 9

  Repeating Patterns

  Module 5 Day 9 Challenge Part 1

  Module 5 Day 9 Challenge Part 2

  Module 5 Day 9 Challenge Part 3

  Module 5 Day 9 Challenge Part 4

  Module 5 Day 9 Challenge Part 5

  Module 5 Day 9 Challenge Part 6

  Module 5 Day 9 Your Turn Part 1

  Module 5 Day 9 Your Turn Part 2

  Module 5 Day 9 Your Turn Part 3

  Module 5 Day 9 Your Turn Part 4

  Module 5 Day 9 Your Turn Part 5
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  How to proof that the rightmost numbers do not repeat if the numbers are in the same group?

  -Define the numbers are in a group if the numbers do not start in the same position (left, middle, and right). (For example, if there are 4 columns, the first 10 rows of numbers are in one group. If there are 5 columns, the first 4 rows of numbers are in one group.)

• Day 10

  Chinese Remainder Theorem

  Module 5 Day 10 Challenge Part 1

  Module 5 Day 10 Challenge Part 2

  Module 5 Day 10 Challenge Part 3

  Module 5 Day 10 Challenge Part 4

  Module 5 Day 10 Challenge Part 5

  Module 5 Day 10 Challenge Part 6

  Module 5 Day 10 Challenge Part 7

  Module 5 Day 10 Your Turn Part 1

  Module 5 Day 10 Your Turn Part 2

  Module 5 Day 10 Your Turn Part 3

  Module 5 Day 10 Your Turn Part 4
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  @neatlobster Great question! Think about the grid box corresponding to a number N. The box has a purple number (N mod 5) and a yellow number (N mod 6).

  We want the yellow number to be even. Write N as 6a + b, where b is N mod 6. For instance, if N is 19, we'd let a = 3 and b = 1. But because any multiple of 6 is even, this implies that 6a is always even. And we want the yellow number, which is b, to be even. So, 6a + b = N must be even since both 6a and b are even. (Basically, the whole point of the bolded sentence is to show that N is even)

  From there, we want the purple number (N mod 5) to be even. BUT we already established that N is even! Therefore, N = 5*(even number) + (even number). (If N = 5*(odd number) + (even number), N would be odd, which can't be true)

  Hope this explanation helped!

• Day 11

  Not Relatively Prime

  Module 5 Day 11 Challenge Part 1

  Module 5 Day 11 Challenge Part 2

  Module 5 Day 11 Challenge Part 3

  Module 5 Day 11 Challenge Part 4

  Module 5 Day 11 Challenge Part 5

  Module 5 Day 11 Your Turn Part 1

  Module 5 Day 11 Your Turn Part 2
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  Thanks

• Day 12

  More Remainders

  Module 5 Day 12 Challenge Part 1

  Module 5 Day 12 Challenge Part 2

  Module 5 Day 12 Challenge Part 3

  Module 5 Day 12 Challenge Part 4

  Module 5 Day 12 Challenge Part 5

  Module 5 Day 12 Your Turn Part 1

  Module 5 Day 12 Your Turn Part 2

  Module 5 Day 12 Your Turn Part 3

  Module 5 Day 12 Your Turn Part 4

  Module 5 Day 12 Your Turn Part 5
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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).