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Today at my lecture the teacher taught following thing about remainders:-

He started of explaining with the help of a question that is to find the no. of possible remainders and the values of those remainders as well when 45^n is divided with 63

He started off like this :-

  1. Possible remainders with 63 = {0,1,2,....62}
  2. Now since HCF(45,63)=9 ; the remainders have to be multiple of 9, so our new list of remainders would now be {0,9,18,27,36,45,54}
  3. Also since prime factor of 7 is not present in numerator , the remainders wont be a multiple of 7 , so the final revised list of remainders would be {9,18,27,36,45,54} and then he further went to on to show that after the cycle of every 6th power of 45^n we will again repeat with same remainders ; like 45^n will give same remainder as 45^(6+n)

Now to test this concept, I took the question on my own as to find out the values and number of remainders possible for 9^n/12 ;

  1. So possible remainders with 12= {0,1,2,....,11}
  2. HCF of (12,9) =3 ; so the possible set of remainders will be {0,3,6,9}
  3. Now the prime factor 2 is not present in numerator , therefore removing its multiples

The final set of remainders come out to be {3,9}; but upon checking for few values like 9^1 , 9^2 , I am always getting 9 as remainder and not 3

Please let me know why the algorithm is not working here

Fin27
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    You have been around for more than a year. Haven't you yet noticed that you are supposed to use MathJax around here? – José Carlos Santos Dec 29 '22 at 18:09
  • "since prime factor __ is not present, the remainders won't be..." That is true, and so the list you are left with will be a superset of the list of all possible remainders. It could be that everything in the list is possible, but it could also be that there still remain some value(s) which still need to be removed. – JMoravitz Dec 29 '22 at 18:13
  • @JMoravitz, thank you ! , why is that some values have to be removed? When and what check should one finally put on the last list we get by removing the prime factors that are not present in the Numerator? – Fin27 Dec 29 '22 at 18:19
  • An easy way to do such problems is to use the mod distributive law

    $$\bmod cn!:\ ca \equiv c(a\bmod n)\qquad$$

    Thus $\bmod 12!:,\ 9^k = 3(3\cdot \color{#c00}9^{k-1}\bmod 4) = 3(3\cdot \color{#c00}1^{n-1}\bmod 4) = 3(3),,$ for $,k\ge 1\ \ $

     – Bill Dubuque Dec 29 '22 at 19:05
  • @BillDubuque Step 3 :- Since 7 is a prime number which is only present in the prime factorization of denominator and not the numerator , the remainder after division has to be a "non-multiple" of it – Fin27 Dec 29 '22 at 20:06
  • We need to work mod $4$ (not mod $2)$, i.e. $!\bmod 4!:\ 9^n\equiv 1^n\equiv 1.,$ Among $,0,3,6,9.,$ only $,9\equiv 1\pmod{!4}.,$ This is equivalent to using CRT to solve the system $,x\equiv 0\pmod{3},\ x\equiv 1\pmod{4}.,$ But you worked mod $3$ and mod $2$ which only determines the result mod $6.\ \ $ – Bill Dubuque Dec 29 '22 at 20:49

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