rsrighosh wrote:
I read everybody's comments here, but I am unable to understand how to proceed. I got so many doubts now
a) how everyone reached a conclusion.... remainder cyclicity of 2 is 3..
i) I realised that 51 mod 7 is 2, \(51^2\) mod 7 is 4.... bit it is difficult and time consuming to find \(51^3\) mod 7. Although I still did that to try to understand the concept however i saw \(51^3\) mod 7 is 1 (which is \(2^0\)). Does this mean every time in such divisions (\(\frac{a^n}{b}\)), "1" as a remainder is common?
ii) After this I had to use calculator to find 51^4 to understand that the value of 51^4 mod 7 is again 2... which is repeating. So is there any easier alternate method to find remainder cyclicity
b) What was the logic behind concluding we need to do this:- \(\frac{2^{203}}{7} \)---> \(8^67 \) x \((2)\) x \((2)\)
How did everyone reached a conclusion that we have to use 8 here.. is it because 8>7 and remainder cannot be greater than 7?
Is there any online post which can clear the concept of such kind of concepts of remainders?
Today after 4 months I found a post on such remainder question. So here to answering myself
\(\frac{51^{203}}{7}\) can be written as \(\frac{(49+2)^{203}}{7}\)
Now with binomial concept we will be left with \(\frac{2^{203}}{7}\) while other terms will all be divisible by 7.
Now checking the remainder cyclicity
\(2 mod 7 = 2\) -----> 3k+1
\(2^2 mod 7 = 4\) --> 3k+2
\(2^3 mod 7 = 1\) --> 3k
Take consecutive powers of 2 (e.g. \(2^1,2^2,2^3\)) until you find the remainder of 1. That will tell you what the cycle is (e.g. it could be q cycle of 3,4,5 etc). Then, once you figure that out, divided the large exponent by that result.
So we see that the cyclicity is \(2^{3k+p}\) ( \(3\))
So, \(\frac{2^{203}}{7}\) can be calculated as follows
\(203 mod 3\) is \(2\)
Hence remainder is \(4\)
Answer: A
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