Hearted Youtube comments on Mathologer (@Mathologer) channel.
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Always, fascinating!
I followed, surprisingly, until the last example. That's when my brain went flat and would inflate no more, until patched with a second cup of tea.
Why, at the end of each video, is there a smile on my face, and a chuckle in my heart? Perhaps they have much to do with your own gentle encouragement along the way with a chuckle here and a giggle there, however, I'm also quite fond of that sweet tune, pleasingly harmonized on a collection of folksy strings, that serves as a segway and ending to your videos.
As always, thank you and I'll look forward to the next time. š
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4:50 already every power of 9 plus one is special, because the triangles can be collapsed in like so: every 9th hexagon is sufficient to account for 9 generations in the future, so ignore the eight between that and the previous one. 10 is good, and a group of 82 would behave just like a group of 10 (with 8 between each one, 8*9 = 72, which is 82-10), and generation 10 of 82 hexagons would behave just like generation 1 of 10 hexagons. likewise generation 82 (assuming the first row is generation 1) is determined only by the top corners in the same way. Therefore 82 behaves like 10, and so 9^3 + 1 will behave to 82 the way 82 behaves to 10, and so on. My first expectation is that since there are 9 possible rules that this is the root behind why 9 is so special, but it may be that 4 generations also obey this property (powers of 3, plus 1 instead)
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I like the Pigeons at the Olympiad proof best.
For the challenge at the end of chapter 3:
Because every repeating decimal can be written in the form
(x+y)/10^n + x/10^2n + x/10^3n ...
for some integers x,n,y.
We can factor out x, and we can rewrite:
1/10^n + 1/10^2n + 1/10^3n ...
as 1/(10^n - 1)
Noticing that when n is an integer, 10^n and (10^n)-1 are integers, we have the sum of two rationals,
y/10^n + x/(10^n - 1)
which is therefore rational, QED.
For the challenge at the end of chapter 6:
7 integers from 1 to 100 such that no 2 different collections from that set have the same sum:
{1,2,4,8,16,32,64}
This is easily seen by putting the sum in binary form. Each bit in the sum is set by exactly one number from the collection, so if collection A sets a given bit, disjoint collection B lacks the only element that can set it.
For the think-about-it pause at the beginning of chapter 7, I found a much harder and clunkier solution:
There are 2*3*4 = 48 permutation of 4 cards. That's almost enough to pick out 1 card among 52 - if only somebody removed 4 cards from the deck. But somebody did! The mystery card can't be any of the 4 cards you passed to your assistant, which leaves 48.
So we place some natural ordering on the 4 cards. Say, suit is the primary sort key and number/jack/queen/king is the secondary.
We also number the permutations of 4 items - so we have a table that gives us a bijection from a permutation like C D A B to/from a unique index.
And we number the remaining cards in the deck. Same idea: force an ordering, make a table.
Then we look up that card in the deck table, look up that index in the permutations, and arrange the cards in that permutation.
Our assistant does the reverse: Sees what permutation we handed him, looks up its index, constructs a deck table for the ordered deck minus the 4 cards he sees, looks up the permutation index in that, and announces that card as his answer.
Ta daah, except that we need to build large tables on the fly. On the plus side, we can do it for any mystery card, we don't have to choose it ourselves from among 5 cards.
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