Two from the Duke Math Meet
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9:15 man's room is falling apart!
And fearlessly continuing on without acknowledging it.
Hilariously, the first problem (at least how it is written on the board here) never specifies that the a_i's can't be (pairwise) the same. So the correct answer would be, that there is no maximal natural number m.
By that argument, the second question doesn't have an answer either. He does not specify that m and n have to be natural numbers.
@@programmingpi314 That's true. The answer should be 0 in that case, though.
New proof method: Proof by ambiguity.
@@Ravi-ng3ee Proof π is not a transcendental number: Well, the problem never specified that π is the circle constant 3.1415... so we can just assume π is a variable, we can then assign π=1 to it and see that π is NOT a transcendental number
@@sebastiangudino9377 Now that's funny! Comments like this are why I read comments!
Looks like a game of "19" (instead of "21"), where you have a "3" and some "6"s, while trying to avoid taking more cards than 19.
At the end of the last problem, it would save a bit of time if you went through each value of k and then counted the number of possible n values. For example, setting k=0, we get m=n+3. The highest value that n can be here is 16 since m=16+3=19. So, including 0, n has 17 choices when k=0. Similarly, n has 11 choices and 5 choices when k=1 and k=2, respectively. If k>2, then m>20.
I think 8 minutes were a bit long for proving the result of the first trivial problem, since it's easy to show that for any n >= 0 the 5 consecutive factorials (5n)!, (5n+1)!, (5n+2)!, (5n+3)! and (5n+4)! always have the same number of trailing zeros, and that (5n+5)! always has at least one more trailing zero than (5n)! And that's a good place to stop.
Just to be safe I'd prove that (5n-1)! has fewer 0s.
I handled the 2nd problem a little differently. I first used the fact that 2^k mod(9) takes on the pattern 1, 2, 4, -1, -2, -4, ... This makes it obvious that 1 pairs with -1, 2 pairs with -2, and 4 pairs with -4 to get 0 mod(9).
Second, I simply made a 20x20 table going from 0 to 19. The columns represent 2^0, 2^1, 2^2, ... which is 1, 2, 4, ... mod(9) respectively. Same idea for the rows. This table will have 4 columns and rows corresponding to each of 1 mod(9) and 2 mod(9). There will be 3 columns and 3 rows corresponding to each of 4 mod(9), -1 mod(9), -2 mod(9), and -4 mod(9).
The probability we're interested in is just 2*P(picking from a 1 mod(9) column and -1 mod(9) row) + 2*P(picking from a 2 mod(9) column and -2 mod(9) row) + 2*P(picking from a 4 mod(9) column and -4 mod(9) row). Note, my way needs to multiply by 2 since I didn't restrict the probability space by asserting n
I got the same answer, with the note that this assumes n and m are restricted to nonnegative integers. If they can be negative, then there's infinitely many noninteger results possible, e.g. 2^-3 + 2^4 is 16⅛, which is in the allowed range. Since there are infinitely many of these and only finitely many congruent mod 9, the probability is actually 0.
Figuring out how many zeros n! has at the end was one of my favorite problems as an undergrad
Also the answer is no factorial meets this criterion (in base 1; no base was defined). :>
We may consider "base 1" as the natural base of numbers since it is the only representation of the numbers such that quantity is preserved by the representation itself with no further rule.
If we consider what number and base is, taken in the usual and literal sense would require every digit to be a coefficient of 0, but this would destroy the notion of a base as a base exists to represent comparabilities [numbers from now on], so it must be something else in order to support numbers. We can instead reformulate the numbers as the conjunction of the smallest abstract subsistence [subsistence from now on] that, on its own, is not comparable, yet with another of itself, becomes comparable (and enumerable), as water is wetness, yet is not wet because to be wet is to have water upon a material body, but water is a material body, therefore a water molecule in contact with one or more water molecules is wet, so also a mere subsistence is not comparable (enumerable) with itself, and nothingness is not a subsistence, so what remains is that the smallest number is 11, yet it relies on the first subsistence to subsist, and yet we must also recognize that 11 is 1 more than 1, and that 1 is 1 less than 11, so we must recognize that 1 in the presence of numbers itself becomes a number; and in their absence, is a subsistence. This makes a mathematical formulation of the numbers that is more concrete, less abstract, bringing it more in line with traditional mathematics, if not equal to them.
Base 2 then looks no different than before: 1, 2, 11, 12, 21, etc., the carry rules unchanged, only the symbols different (but choice of symbol of course doesn't matter).
Base 3: 1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, etc.
...
Base N: 1, ... N, 11, ... NN, 111, ... NNN, ... etc.
I wish programmers would recognize the pidgeonhole principle with respect to multithreading, and that the designers of the Windows NT and Linux kernels would've in fact done so as well 😔
12:30 The problem is just very slightly simpler to see if you instead use the corresponding negative numbers in the mod 9 table, i
E. it reads 1 , 2, 4, -1, -2, -4, 1. That way it’s really obvious you are adding a positive and a negative together which add to 0 mod 9.
Also if you think of cycling through that list up to 20, you have 4 1s, 4 2s, 3 each of 4,-1,-2,-4 ... so unique sums to 0 are 12+12+9=33 pretty immediately.
Are you planning on making any videos about PDEs?
I read the title of the video as "a couple of problems from Duke Math get to know one another"
Fascinating!
Merci
18:43
9:09 Rewound this thinking it was an editing artefact whose point I’d missed. Nope, just a ‘set malfunction’.
You forgot the zero in the thumbnail.
Hi,
8:23 And what is the question?