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[preface, i'll use '%' as the modulo or remainder operator, as it is in C and many other programming languages, where for example 10 % 7 == 3]

i'm looking for an algorithm to solve for n in

(A*n + B) % C == 0

where A, B, and C are all positive integers.

i have additional constraints on my particular problem that i don't think are important for the solution. (namely, A is less than C, A is a power of 3, C is a power of 2, and B is odd, all of which i believe imply that n must be odd.) (and yes, i'm toying with Collatz here)

i can brute force this (try all possible n from 0 to C-1), but that will become time consuming as C increases. i could also try to binary search for it, which would be a performance improvement, but i'm looking for a more efficient algorithm.

Gorilla Sapiens
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  • You'll find some code optimization here, and code in the comments/chat: https://math.stackexchange.com/questions/3454674/calculate-the-maximum-in-the-collatz-sequence – Collag3n Apr 17 '21 at 21:20
  • See the linked dupes for the standard theory and algorithms for solving linear congruences. – Bill Dubuque Apr 17 '21 at 22:08

1 Answers1

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Translating to more standard number theory, you want a solution to $$ An\equiv-B\pmod C $$ The first step to a solution is to solve $$ Am\equiv 1\pmod C $$ We then get $n=-Bm$ by multiplying both sides with $-B$.

The extended Euclidean algorithm gives you two integers $m, m'$ such that $$ Am+Cm'=1 $$ (as your additional constraints tell us that $\gcd(A,C)=1$). This finishes the solution. A solution between $0$ and $C$ is not guaranteed in this case, so taking $n\%C$ at the end will give you that.

Arthur
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  • But the first step fails if $A$ and $C$ are not coprime. But not to worry since this is FAQ which has been answered hundreds of times in the past. Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Apr 17 '21 at 22:10
  • @BillDubuque This time I actually did a rudimentary search for dupes and still didn't find any good ones. Please stop hounding me about it, I have decided it is impossible. (I'm sure some can do it, but I just can't.) – Arthur Apr 18 '21 at 04:13
  • so... any clues on how to do the first step (Am == 1 (mod C) ??? – Gorilla Sapiens Apr 18 '21 at 21:57
  • @GorillaSapiens Yes, the extended Euclidean algorithm, as I said. – Arthur Apr 19 '21 at 07:08