the most creative digit sum problem I have seen!! 

The only other solutions seem to be 518 and 598.

8:38 the easier way to show that one of those terms is divisible by 11 is to notice that 11 is prime, and apply the Fundamental Theorem of Arithmetic.

Note: 0

The three digit problem has four 'proper' solutions (135, 175, 518, 598) along with three improper solutions (1, 43, 63). Extending the problem to four digits with the sum being f(n=abcd) = a+b^2+c^3+c^4 gives three proper solutions (1306, 1676, 2427) along with the trivial improper solution of 1. How many solutions might there be for the equivalent five digit problem?

9:13 Missed the case that c-1 = 0 here. That also doesn't work because it leads to N=001.

Hey, do you mind finding the value of cos 4 degrees?

If we don't limit ourselves to 3 digit nos (i.e., require a>0 in Penn's notation), then 1, 43, and 63 are also solutions.

You can do it a bit less rigorously just by listing cases. Solutions are 135, 175, 518 and 598 a + b^2 + c^3 = 100a + 10b + c a=1 to 9; b,c = 0 to 9 99a + b(10 - b) + c(1 - c^2) = 0 b 0 1 2 3 4 5 6 7 8 9 b(10-b) 0 9 16 21 24 25 24 21 16 9 c (c-1)c(c+1) = 99a + b(10-b) 0 0 1 0 2 6 3 24 4 60 5 120 = 99 + 21, abc = 135, 175 6 210 = 99 + 111, a=1, b(10-b)=111=3x37, no solution = 198 + 12, a=2, b(10-b)=12, no solution 7 336 = 99 + 237, no solution = 198 + 134, no sol = 297 + 39, no sol 8 504= 495 + 9, abc = 518, 598 9 720 = 693 + 27, no sol

9:37 you have to make the same argument for c=1!!!

thanks

Here's the list of solutions for digit count up to 8 (interestingly, there are no solutions for 5, 6, or 8 digit numbers): 89 135 175 518 598 1306 1676 2427 2646798 It would be interesting to categorize which digit counts do have solutions.

for a in range(100,1000): x = (a//100) y = (a - x*100)//10 z = (a - x*100 - y*10) if a == (x + y**2 + z**3): print(a)

Possible solutions: 000, 001, 043, 063, 135, 175, 518, 598

You could also simply spend 20 seconds to write a small script to find the solutions 😄

In my opinion is simpler to check which b and c values satisfy b(b-10) mod 99 = c(c^2-1) mod 99, and then compute a

there has to be a better approach than just listing cases. I do think it'll boil down to listing cases, but i think we should be able to do some analysis to reduce the no. of cases to check.

Since we found the fixed points of the function let's move on to finding its other periodic points :) Starting from f(f(N))=N

9:45, what??! Did he say "B times 10 B minus 1"??!

What about for number of digits larger than 3?

Hi Michael, I've been watching your channel for a while and I've big fan of your videos! I have a problem suggestion that I thik could make for a good video idea. My Linear Algebra teacher had a couple of exercises about matrices and linear systems reduced mod p. 3 years later and I've never been able to solve one of these whch asked how many m by n matrices whose entries where reduced mod 2 had rank 1. I'm not sure if this is how you usually take problem suggestions, or if the problem is actually simple and I'm just being dumb, but I think the whole idea would make for a nice video.

First reaction is, why do the powers in that manner (left to right)? This forces knowing in advance the number of digits of the source number. Doing it right to left would create one single function that works for source numbers regardless of their number of digits... Weird

An easy method readily springs to mind: write a program to test the cases. There's only 900, so you needn't even spend time looking for a subtle way to save time. Programming is skill not valued highly enough by mathematicians. If the person who composed the question didn't intend to test the pupils' proficiency in that skill, they shouldn't have composed a question like that.

First :*