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Cake day: June 19th, 2023

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  • It works differently for each number. For 2, the last number has to be divisible by 2. For 3, the sum of the digits has to be divisible by 3 For 5, the number has to end with a 0 or a 5. For 7, it is kinda tricky. Take the last digit, double it, and subtract it from the numbers on the left. If the remainder is 0 or divisible by 7, the whole number is divisible by 7. For example 49: 9×2=18, 4-18=-14, -14/7=2 with remainder 0. For 700, 0×2=0, 70-0=70, 70/7=10 remainder 0.

    This is usually specified for prime numbers, for non-prime number, you just do calculate the prime components of a number and combine the rules.

    For example, divisibility by 15: it has to be divisible by 3 and 5. 1+5=6, 6/3=2 remainder 0. 15 ends with a 5. For number where with multiple same prime components the rules for these duplicate numbers have to apply multiple times. Like for 25, it has to end with a 5 or 0, and when dividing the number by 5, the result has to end with a 5 or a 0 aswell.














  • derpgon@programming.devtoGaming@lemmy.mlClassic Microsoft
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    5 months ago

    Also swapped performance issues with more bugs that were there years ago and still persist because they are almost impossible to fix.

    In the end, we all know it was done either because they wanted to shove microtransactions down our throat and/or had some kind of deal to maintain the Java edition without microtransactions.

    Or, maybe, just maybe, they though “it’s just a block game, what could be so hard to rewrite it?” and absolutely failed what a single person got right almost immediately back in the day (like not falling through the floor all the time).