Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. Challenges: 1) We can guarantee at least 5 on one side. There are two ways to split the pigeons: 3-2 and 4-1. The max you can hit for both of them is 3 and 4 respectively. So we have at least 3 pigeons on one side. Add in the two pigeons on the equator and we have 5. 2) let the mystery value equal x. by multiplying by 10^some integer until the digits line up again, and then subtracting, we have 10^whatever*x - x = the non-repeating part you're left with. Factor out x on the left and that 10^whatever - 1 becomes a bumch of 9s. Then divide and we have the non-repeating remainder over a bunch of 9s. If ever the remainder isn't an integer just multiply both the numerator and denominator by 10 to knock out the decimal. Example: 0.4318181818... x = 0.4318181818... 100x = 43.18181818.... 99x = 42.75 (all the 18s afterward cancel out) x = 42.75/99 x = 4275/9900 x = 19/44 (simplify) 3) let's pretend we color the "didn't shake hands" white. We leave the "shook hands" black. So all we have to do is prove a triangle exists somewhere, and it doesn't matter what color. Note that each dot will always have at least three lines of the same color connecting to it, for the same reason the pigeons in challenge 1 can be split into a side with at least 3 and a side with less. Look at a point, and follow three lines of the same color (I'll choose black) to three other points. Any line connecting two of these three points would either be black or white. If it were black, that would make a triangle with the two points and the starting point. If it weren't, then the three dots all have white lines between them, which makes another different triangle. 4) 315. And no, I didn't count manually. Here's how I found out: The 7 edges UB, UR, RB, DB, DL, LB, and UL, all commute in one big cycle of length 7. The remaining 5 edges do the same. The 5 corners URF, ULF, DRF, DLF, and DRB go back in their positions but in the wrong orientation. So it takes 3 cycles to get back to their original orientation. The same thing happens with the other 3 corners in the back, only this time they take 9 cycles because they permute around each other too as well as reorient themselves. All that's left is to compute the lcm of 7, 5, 3, and 9, which turns out to be 315. 5) 1, 2, 4, 8, 16, 32, and 64. Any collection just results in the binary representation of the number. Since every number can be written in binary in only one way, every single collection will add to a different number. 6) Queen of Hearts. The 9 of Hearts at the start signals the suit of the missing card is Hearts. The remaining three kings can be sorted as MBT (Diamond, Club, Spade), which is assigned 3. 3 spots after 9 is the Queen, and putting it all together we have Queen of Hearts. Edit: I forgot one, challenge 5 at chapter 6
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