is beehaw related to lemmy?

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Joined 1 year ago
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Cake day: June 8th, 2023

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  • Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

    For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + … + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you’re left with n = a_0 + a_1 + … + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + … + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3





  • That’s some high IQ usage of a meme. Lemme see if I’m getting this right:

    • the total area of the image ( = RHS of the equation) is 1
    • you divide the image into 4 parts so that the area of 1 part is is 1/4 ( = 1/2^(2*1)). You take the first three quarters and leave the fourth quarter for recursion (I’ll call it x1). That gives you 3(1/4) + x1 = 1
    • now you take x1 and do the same with it. This time, the area of each sub-quarter is 1/16 ( = 1/2^(2*2)). Three such sub-quarters and a leftover x2 gives you 3(1/16) + x2 = x1. Put this back into the first equation to get 3(1/4 + 1/16) + x2 = 1.
    • repeat until infinity; each time the area of the resulting tile is 1/4 of the previous tile (which is the 2n in the exponent part)

    Edit: imma remove all markdown since it doesn’t seem to work, at least on liftoff. Enjoy the lisp-like mess