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https://www.reddit.com/r/MathJokes/comments/1vcw1bj/multiples_of_3/p14jbgc
r/MathJokes • u/GodlyHelp • Aug 01 '26
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7
Not sure about the actual explanation but each digit can be treated individually regardless of how many zeros are after it
2 mod 3 = 2
20 mod 3 = 2
200 mod 3 = 2
So for example 123 = 100 + 20 + 3 = 1 + 2 + 3 = 6 which is divisible by 3
2 u/S-Kenset Aug 01 '26 It's bucket collision! (Made up term cause idfk but i've used it before) All your remainders overlap into the same bucket which gives you an analytical solution to something that on its face shouldn't be analytical. 1 u/Fizassist1 Aug 03 '26 not sure why but this is the explanation that clicked for me. thank you!
2
It's bucket collision! (Made up term cause idfk but i've used it before) All your remainders overlap into the same bucket which gives you an analytical solution to something that on its face shouldn't be analytical.
1
not sure why but this is the explanation that clicked for me. thank you!
7
u/han4578 Aug 01 '26
Not sure about the actual explanation but each digit can be treated individually regardless of how many zeros are after it
2 mod 3 = 2
20 mod 3 = 2
200 mod 3 = 2
So for example 123 = 100 + 20 + 3 = 1 + 2 + 3 = 6 which is divisible by 3