The Blog I Neglected (But I'm Back!)

By Brownie -

Admittedly, I’ve let this blog gather dust for quite some time. Life happens, right? But sometimes, all it takes is a surprise piece of mail to bring you back.

Enter: a 1099-NEC in my mailbox.

For context: years ago, I had an AdSense account—one I hadn’t touched in a decade. Turns out, my account was deactivated for being inactive, and I never reached the minimum payment threshold. But here’s the kicker: Google still sent me a 1099-NEC, complete with a $3 federal tax withholding. Where did the rest of the money go? Apparently, it was sent to my state’s unclaimed funds. Big tech at its finest.

Oh, and just to add a layer of nostalgia: at some point, I couldn’t log into AdSense because I had used Google+ single sign-on. Remember Google+? Yeah, me neither.

But I digress.

Now, I’m here to revive this blog and make it a small corner of the internet for solving math problems, maybe once a month. There’s something refreshing about the simplicity of old-school blogging—a little less noise, a little more focus.

So here’s to fresh starts, solving puzzles, and maybe even earning enough to hit that payment threshold this time around. (JK)

Math Puzzle of the Day

The 1099-NEC I received was issued by “XXVI Holdings” in Mountain View, CA.

Here’s the puzzle:

  • XXVI = 26
  • 26 = the number of letters in the English alphabet.

AI-Generated Visual Flair

To top it all off, I’ve included an AI-generated image of a 1099-NEC to give this blog post an extra visual flair.

AI-generated 1099-NEC tax form

Comments

Post a Comment

Popular posts from this blog

Project Euler Problem 684 Inverse Digit Sum - Solution

By Brownie -
Image
This one is relatively straight forward. Below is a solution in python and will give us correct answer within reasonable amount of time. (Generally, if python can calculate this fast enough, I am happy with the solution, nothing against python...) modulo = 1000000007 #speed up hint: (10e9) % modulo = -7 #init speedup array shortcut = 18 arr_lookup = [ 1 ] * (shortcut + 1 ) for i in range ( 1 ,(shortcut + 1 )): arr_lookup[i] = ( 10 ** i) % modulo def S (i): A = i % 9 B = i // 9 #sum = (1+..+A)*(10^B) + A*(10^B-1) + 45*(10^(B-1)) + 9*(10^(B-1)-1) + 45*(10^(B-2)) + 9*(10^(B-2)-1) + ... + 45*10 + 9*9 +45 #sum = (A(A+1)/2)*(10^B) + A*(10^B) - A +(9+45)*(10^(B-1) + .. + 1) - 9*B #sum = (10^B) * (A(A+1)/2 + A) - A - 9*B + 6*(10-1)*(10^(B-1) + .. + 1) = (10^B) * (A(A+1)/2 + A) - A - 9*B + 6*(10^B -1) #sum = (10^B) * (A(A+1)/2 + A + 6) - (A + 9*B + 6) #since (7,10e9+7)=1; 7^(10e9+7 - 1)%(10^9+7)=1 #B = 18C+R C = B ...
Read more

Project Euler Problem 686 Powers of Two - Solution

By Brownie -
Image
Powers of Two     Problem 686 2 7 = 128 2 7 = 128  is the first power of two whose leading digits are "12". The next power of two whose leading digits are "12" is  2 80 2 80 . Define  p ( L , n ) p ( L , n )  to be the  n n th-smallest value of  j j  such that the base 10 representation of  2 j 2 j  begins with the digits of  L L . So  p ( 12 , 1 ) = 7 p ( 12 , 1 ) = 7  and  p ( 12 , 2 ) = 80 p ( 12 , 2 ) = 80 . You are also given that  p ( 123 , 45 ) = 12710 p ( 123 , 45 ) = 12710 . Find  p ( 123 , 678910 ) p ( 123 , 678910 ) . Test the first 3 digits of a large number is quite costly.  So first thing first, think log(). --------------------- import math lowerbound = math.log(1.23,10) upperbound = math.log(1.24,10) def testlog(i): logi = math.log(2,10)*i diff = (logi - int(logi)) return diff We only care about the part after decimal point, thus  dif...
Read more

Project Euler Problem 717 Summation of a Modular Formula - Solution

By Brownie -
Image
Summation of a Modular Formula          Problem 717 For an odd prime  p p , define  f ( p ) = ⌊ 2 ( 2 p ) p ⌋ mod 2 p f ( p ) = ⌊ 2 ( 2 p ) p ⌋ mod 2 p For example, when  p = 3 p = 3 ,  ⌊ 2 8 / 3 ⌋ = 85 ≡ 5 ( mod 8 ) ⌊ 2 8 / 3 ⌋ = 85 ≡ 5 ( mod 8 )  and so  f ( 3 ) = 5 f ( 3 ) = 5 . Further define  g ( p ) = f ( p ) mod p g ( p ) = f ( p ) mod p . You are given  g ( 31 ) = 17 g ( 31 ) = 17 . Now define  G ( N ) G ( N )  to be the summation of  g ( p ) g ( p )  for all odd primes less than  N N . You are given  G ( 100 ) = 474 G ( 100 ) = 474  and  G ( 10 4 ) = 2819236 G ( 10 4 ) = 2819236 . Find  G ( 10 7 ) This problem is not really that complicated to reduce to acceptable execution time. (Python, ~8hrs in 1 thread on a typical laptop.) Please leave a comment if you have better solution or suggestion. See note below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...
Read more