Subcategories

• Module 5 Day 3 Challenge Part 1

  Divisibility by 12

  0
  Topics
  0
  Posts
  No new posts.
• Module 5 Day 3 Challenge Part 2

  Divisibility by 12

  1
  Topics
  1
  Posts

  

  The forum link after the quiz leads to the quiz, not the forum.

• Module 5 Day 3 Challenge Part 3

  Remainders Mod 2

  1
  Topics
  3
  Posts

  

  I really love the symmetry here! 👏 👏 ☯

  This is a great way! And the minus signs all cancel out very nicely.

  You asked for another way, and here is a slightly different way which really isn't all that different from the original method, but it simplifies some of the intermediate calculation a little bit. 🔨 You can multiply the first few together, and then find the remainder of that product at each step before multiplying the other numbers.

  $$ 71 \text{ is } 1 \text{ mod } 7 $$

  $$ 72 \text{ is } 2 \text{ mod } 7 $$

  $$ 73 \text{ is } 3 \text{ mod } 7 $$

  $$ \rightarrow 71 \times 72 \times 73 = 1 \times 2 \times 3 = \textcolor{red}{-1 \text{ mod } 7} $$

  Now, let's multiply the other numbers!

  $$ 74 \text{ is } 4 \text{ mod } 7 $$

  $$ \rightarrow \textcolor{blue}{ 71 \times 72 \times 73 \times 74 } = -1 \times 4 = -4 = -4 + 7 = \textcolor{red}{3 \text{ mod } 7} $$

  Now let's multiply the \(75.\)

  $$ 75 \text{ is } 5 \text{ mod } 7 $$

  $$ \rightarrow \textcolor{blue}{71 \times 72 \times 73 \times 74 \times 75} = 3 \times 5 = 14 + 1 = \textcolor{red}{1 \text{ mod } 7} $$

  And finally, we've got the \(76.\)

  $$ 76 \text{ is } 6 \text{ mod } 7 $$

  $$ ( 1 \text{ mod } 7 ) (6 \text{ mod } 7 ) = \boxed{ \textcolor{red}{6 \text{ mod } 7 }}. $$

  This way, your "running total of your remainder" is always less than \(7,\) so there are just smaller numbers to deal with. The are so many little shortcuts to calculating this, and it's kind of like a puzzle to find them! 🎉

• Module 5 Day 3 Challenge Part 4

  Trick for Divisibility by 4

  2
  Topics
  4
  Posts

  

  @amusingminnow Great question! So, by definition, composite numbers are numbers that have more than \(2\) factors, and prime numbers are numbers that have exactly \(2\) factors: \(1\) and itself. The thing with \(1\) is that it only has one factor: \(1!\) so, it's neither prime nor composite.

  \(1\) is just a unit, so it's called a unit sometimes.

• Module 5 Day 3 Your Turn Part 1

  Trick for Divisibility by 8

  1
  Topics
  12
  Posts

  

  Cool $$$$

• Module 5 Day 3 Your Turn Part 2

  Eliminating Choices

  0
  Topics
  0
  Posts
  No new posts.
• Module 5 Day 3 Your Turn Part 3

  Trick Using Negative Remainders

  2
  Topics
  11
  Posts

  

  @amusingminnow Yeah! You can pick any modulo you want for these types of tests. Although, he reason why Prof. Loh chose mod \(9\) is because it's really easy to find both \(8^8 \pmod{9}\) and the really huge numbers\(\pmod{9}\), since (digit sum of number) mod 9 = (number) mod 9. Other modulos might not be as convenient.

  If you wanted to use\(\pmod{3}\), it would look something like this:
  since \(8 \equiv 2 \equiv -1 \pmod{3}\),
  \(8^8\pmod{3}\equiv(-1)^8\pmod{3}\equiv 1\pmod{3}\).

  So, out of the answer choices, we have to find the number that's also \(1\pmod{3}\).
  The thing is, digit sums actually works for mod 3 too, it's the exact same process. Doing so will give you that \(16777216\equiv1\pmod{3}\). Same answer!

  \(3\) and \(9\) are nice because we can find big numbers mod 3 and mod 9 really easily. For numbers like \(7\) though, it might be tougher (if you want to, go ahead and try it out!)

  For this problem though, I think mod 9 is best. Consider the fact that

  \(1,4,7 \equiv 1\pmod{3}\), but they are \(\equiv1,4,7\pmod{9}\), respectively. Luckily, for this problem, we didn't run into any confusion, so mod 3 worked fine. However, other problems might be different, and the fact that mod 9 is larger can help in some situations. Hope this makes sense, and let me know if you have any other questions! 🙂

• Module 5 Day 3 Your Turn Part 4

  Trick Using Negative Remainders

  0
  Topics
  0
  Posts
  No new posts.