Just a quick question:

I'm trying to find the least residue of 7^5 mod 50.

I've been messing about with Chinese remainder theorem etc... But it seems a bit easy which is throwing me off and making me think I'm solving another question.

So instinctively I know that 7^2=49 so 49+1=50. Does that mean 7 is the least residue?? As in 7x7 = 49 +1 = 50?

Sorry this is very basic but this is my first go at modular arithmetic (well knowingly).

Thanks

I think you are having difficulty understanding the meaning of modular arithmetic

I will use an example
e.g. what is 15 mod 12
Think of a clock, after we have gone around counting the hours 1,2,3,... etc we start over again after 12 hours
so it becomes obvious that 15mod12 = 3
or
think of it this way:
What is the remainder when I divide 15 by 12 ?

so to our question:
7^5 mod 50
= 16807 mod 50 ---> 16807/50 = 336.14
so could take away 336(50) from 16807 to leave us with 16807 - 336(50) = 7

thus: 7^5 mod 50 = 7
=

Well op i feel like you might be doing the same TMA as me :'D

basically what I did is break down 7^5 into (7^2)*(7^3)

I then worked that out and -50

I got 1 and 13... obviously 1 is the least residual

No problem, I'm here to help! Finding the least residue of 7^5 mod 50 involves using modular arithmetic.

To calculate the least residue, you need to take the remainder when 7^5 is divided by 50.

However, instead of calculating each term of the power separately, it's more efficient to use the concept of modular exponentiation. This involves repeatedly squaring the base and reducing modulo 50.

Here's how you can do it step by step:

1. Start with 7 as the base.
2. Square the base: 7^2 = 49.
3. Reduce modulo 50: 49 mod 50 = 49.
4. Square the result again: 49^2 = 2401.
5. Reduce modulo 50: 2401 mod 50 = 1.
6. Now, raise the result to the power of 5. We could repeat the squaring process four more times, but since we already found 49^2 = 1 (mod 50), we can simply multiply 1 by itself four more times.
1^5 = 1.
7. Finally, take the least residue by reducing modulo 50: 1 mod 50 = 1.

Therefore, the least residue of 7^5 mod 50 is 1.