Almost All Numbers Contain the Digit Three

My mind was blown today when I came across the assertion (and proof!) that almost all numbers contain the digit '3'.

For small numbers of a given length of digits, you can count the ones that contain the digit '3'. You can see these values in the following table:

Number of digits	Examples	% Containing 3
1 (0-9)	3	10% (1)
2 (0-99)	3,13,23,30,31,32,33,34,35,36,37,38,39,43,53,...	19% (19)
3 (0-999)	3,13,...,30-39,...,300-399,430-439,...	27% (271)

You can count them by hand, but the count 'C' for any given number of digits 'n' is given by the following formula:

C = 10^n - 9^n

Thus the percentage is given by:

% = (10^n - 9^n) / 10^n
% = (10^n / 10^n) - (9^n / 10^n)
% = 1 - (9/10)^n

Taking the limit of this a n -> inf and you see that % = 1.

James Grimes does an excellent job of explaining it in this video:
https://www.youtube.com/watch?v=UfEiJJGv4CE

Palindrome Dates

If you follow the traditional American format of mm/dd/yyyy for dates, then today, 7/10/2017 is "palindromic". The next date to follow this format will be 8/10/2018. Coincidentally, it also works for the mm/dd/yy format -- 7/10/17. If you follow the European format of dd/mm/yyyy, you're going to have to wait until October 7, 2017 -- which, you guessed it -- also comes out as 7/10/2017. PS. Yes, I know I've played fast and loose with leading zeroes.

Climb To Prime

John H. Conway’s proposed a list of math problems for which he is offering $1000 for a solution.   One of those is his “Climb to a Prime” conjecture, which is stated as:

Problem 5. Climb to a Prime:

Let n be a positive integer. Write the prime factorization in the usual way, e.g. 60 = 2² · 3 · 5, in which the primes are written in increasing order, and exponents of 1 are omitted. Then bring exponents down to the line and omit all multiplication signs, obtaining a number f(n). Now repeat.

So, for example, f(60) = f(2² · 3 · 5) = 2235. Next, because 2235 = 3 · 5 · 149, it maps, under f, to 35149, and since 35149 is prime, it maps to itself. Thus 60 → 2235 → 35149 → 35149 → ..., so we have climbed to a prime, and we stop there forever. The conjecture, in which I seem to be the only believer, is that every number eventually climbs to a prime. The number 20 has not been verified to do so. Observe that 20 → 225 → 3252 → 223271 → ... , eventually getting to more than one hundred digits without yet reaching a prime!

James Davis took up the challenge and found a counterexample: 13532385396179 = 13 ⋅ 53² ⋅ 3853
This prime number fails the conjecture because its factorization maps back to itself!

You can read a few more details here:

http://www.popularmechanics.com/science/math/news/a26815/why-13532385396179-is-a-magic-number/

Finding the Right Math Class

An XKCD strip my kids could all relate to, from the author of "What if..." which they also loved:

https://xkcd.com/1856/