1. Prove that if n is an even positive integer, then n³-4n is always divisible by 48.
2. Prove taht the square of an odd integer is always of the form 8k+1, where k is an integer.
3. Observe that the last two digits of 7² are 49, the last two digits of 7³ are 43, thel ast two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.
Don't know if any are applicable.
The Attempt at a Solution
Basically the homework handed out today for my semester 2 discrete and geometry class in high school. I have no experience in proofs and the teacher assigned this without doing any examples :(
I need help on my form and revision on how I attempt these proofs and basically what I can work on to make them look more convincing and professional.
1. Let n = 48k, 48k+2, 48k+4
n³-4n divisible by 48
Since 192 is divisible by 48, statement is true.
Should I check with 48k+2, 48k+4 afteR?
2. Since k is an odd integer, we can represent it as 2k+1
(2k+1)² = 8k + 1
4k²+4k+1 = 8k +1
8k is divisible by 4k so statement is true.
3. This one i need help the most.
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
and it rotates from 7,9,3,1 (last digit) every 4 numbers...so 7^5 would end with a 7.
so 7^4n+1 would yield a number that ends with 7, 7^4n+2 would yield a number that ends with a 9, etc
so if 7^201 ends with a 7
7^4n+1 = 7^201, n must be an integer
4n+1 = 201
4n = 200
n = 50
Conversely for 7^202
7^4n+1 = 7^202
4n+1 = 202
4n = 201
n is not an integer so 7^202 does not end with a 7.
Getting confused...it looks right sometimes but sometimes it doesn't...am I donig the right stuff?