Bunuel wrote:
gaurav1418z wrote:
Bunuel
(18^22)^10 = 18^220
I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n
so i found remainder for 8^220, and got answer as 1
Where am i going wrong?
Where did I write that? I think that you mean the following part from Number Theory booksaying that
the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4.
Another approach to solve such kind of questions is using cyclicity. Lemme try and explain this.
We have (18^22)^10 = 18^220
Now we try to see what are the remainders for various powers of 18 when it is divided by 7
18^1 leaves 4 as remainder
18^2 leaves 2 as remainder
18^3 leaves 1 as remainder
18^4 leaves 4 as remainder
Hence, we see after 3 set of powers, the remainder starts repeating. Now, this means if the power is multiple of 3, the remainder is 1, if the power leaves remainder 1 when divided by 3, the actual answer will be 4, and so forth.
We see, when 220 is divided by 3, we get 1 as remainder, showcasing 18^1 case, hence the final remainder is 4.
This is because 18^220 = 18^219 x 18 = (18^3)^73 x 18
Thus, we need to find the remainder when 18 is divided by 4 as 18^219 gives 1 as remainder.