It finds the bit at the nth digit and then adds it to a count. It divides by 2 and floors that to make only 1 and 1 return 1. After that, it adds that sum to the nth digit. The math solution is a bunch of if statements being looped.
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This topic is still up and running after like what a months-?
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Ohhhhh I see haha
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I mean… there isn’t a solution to it so yeah.
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It’s been 158 days that a long time -?
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Wow… Just like me doing 100/7 and not figuring it out so I used two pieces of paper doing long division
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bump
Yeah rip. No algebraic solutions. Bitwise operations are everything that runs inside the computer. You’d quite literally have to construct a computer in gimkit to do it without recursion lol.
new project acquired 
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What project? (I’m still curious what that secret project is lol)
I was making a joke about recreating a computer in gimkit, like with actual computer architecture and not just code.
But for the secret project, it’s… secret. It’s also on hold at the moment, I have a lot of stuff going on.
Schoolwork, Librekit, making an app for a school project, being active here, boy scouts… just a ton of stuff to do lol
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Yeah, lots of stuff lately.
question, does anyone know how to replace all 1s in a number with 0?
If so, I might have a solution
Also what are the two numbers-
Is converting them to binary ok?
Set the number to 0.
I mean 45812 to 45802, but with larger numbers and more ones
Any time you need to set all the 1s in a binary number to 0 you get 0 haha
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Well my plan was to add the two numbers (binary form) and remove all 1s and replace with 0, then divide 2.
I just had another idea, but it will only work if the sum of the two numbers (binary form) doesn’t have any zeroes.
You could try adding 10^x to add a one to the beginning of the number, subtract the original number from it, then use a string operation to get all digits from the 2nd letter to the last.
Isn’t that… zero?
Nope, because you add a one to the beginning, then you bypass the issue.
419429 becomes 1000000, then you get it as a string from digit 2 to the end, returning 000000. This way it doesn’t simplify to 0.