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LoginGeneral Tech Bugs & Fixes 3 years ago
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The formulation is naturally a recurrence relation:
f(n)=f(⌊⌊n10⌋/10⌋)+3(n mod 10)+(⌊n10⌋ mod 10)
This, itself, is most naturally viewed mod 100:
f(n)=f(n without its two rightmost digits)+3×rightmost digit+next-rightmost digit
If n has evenly-many digits, then: take n as a base 10 string. Take the first, third, fifth… digits' sum. Take the second, fourth, sixth… digits' sum, and multiply by 3. Add the two together.
If n has odd-many digits, then just stick a 0 on the front and pretend it has evenly-many.
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manpreet
Best Answer
3 years ago
There is some code snippet written on python:
number = 5602004
accum = 0
while number:
accum += (3 * (number % 10))
number = int(number / 10)
accum += (number % 10)
number = int(number / 10)
So, cycle is working while variable number greater then 0. The question is: can this cycle be presented as math formula?