Hi I do not understand the answers to these 2 questions:
1. Can you work out what the remainder is when 2^1000 is divided by 7?
As 1000 = 333 × 3 + 1 then 2^1000 leaves remainder 2 when divided by 7.
2. Expand the brackets of (x − 1)(x − 7) and (x − 3)(x − 5) in mod 8 arithmetic.
(x − 1)(x − 7) = x^2 − 8x + 7 = x^2 − 1 (mod 8);
(x − 3)(x − 5) = x^2 − 8x + 15 = x^2 − 1 (mod 8).
Why is there a "-1", we dont take remainders to be negative?
I'm new to modular arithmetic, just started it right now, so apologies if its a stupid qustion.
Thanks, so we can take 7 or -1, depending on what we need in our answer?
Am i supposed to use this to solve the next question
Without direct calculation, determine the next to last digit in 2^1000? (i.e. the one in the “tens” column.) [Hint: listing the last two digits of each power of 2 you should find a cycle of length 20 which first starts with the second power.]
I don't understand the hint. I've written down the last 2 digits till 2^20, but dont seem to find a pattern. Is this what I was supposed to do?
Since 1000= 333 x 3+ 1, 2^1000= 2^(333 x 3+ 1)= 2((2^3)^333)= 2(8^333). Now 8= 7+ 1 so 8^n= (7+ 1)^n and, by the binomial theorem, that is a sum of multiples of powers of 7 plus 1. 8^n has remainder 1 when divided by 7 for all positive n.
The convention is to reduce all number "mod n" to between 0 and n-1. But, for example, 7, 15, -1 are all equivalent to 7 because they are a multiple of 8 plus 7 (in the case of -1, (-1)8+ 7= -1) but we don't have to. Yes, x^2- 1 is equivalent to x^2+ 7 (mod 8) but we can leave it as x^2- 1 if we want.2. Expand the brackets of (x − 1)(x − 7) and (x − 3)(x − 5) in mod 8 arithmetic.
(x − 1)(x − 7) = x^2 − 8x + 7 = x^2 − 1 (mod 8);
(x − 3)(x − 5) = x^2 − 8x + 15 = x^2 − 1 (mod 8).
Why is there a "-1", we dont take remainders to be negative?
I'm new to modular arithmetic, just started it right now, so apologies if its a stupid qustion.[/QUOTE]