Adding and subtracting different bases with common exponents in ring Z

[morrowcosom]

Homework Statement

Calculate (5^7)-(7^7)+(9^7)-(11^7) in Mod24

Homework Equations

The Attempt at a Solution

I added all the bases and got -4 (which I changed to 4), then I took 4^7 and ended up with 16,384. I divided 16,384 by 24 as many times as I could, which gave me an end result of 40, and being that that mod is 24 my final answer is 16.

I am doing independent study on a computer program and it says 20, what did I do wrong? Maybe it had something to do with how I added the bases?

Thanks

 

[Mark44]

Homework Statement

Calculate (5^7)-(7^7)+(9^7)-(11^7) in Mod24

Homework Equations

The Attempt at a Solution

I added all the bases and got -4 (which I changed to 4), then I took 4^7 and ended up with 16,384. I divided 16,384 by 24 as many times as I could, which gave me an end result of 40, and being that that mod is 24 my final answer is 16.

I am doing independent study on a computer program and it says 20, what did I do wrong? Maybe it had something to do with how I added the bases?

I don't think you want to add the bases. I can't think of any good reason for doing that.

Also, -4 and + 4 aren't in the same equivalence class, so you can't just change one to another. -4 is in the same equivalence class as 20, in modulo 24.

Let's take these powers in small steps, starting with 5⁷.
5 $\equiv$ 5 (mod 24)
5² = 25 $\equiv$ 1 (mod 24)

So 5⁷ = 5 * 5⁶ = 5 * (5²)³ $\equiv$ 5 * 1³ (mod 24) = 5 (mod 24).

If you divide 5⁷ by 24, you get a remainder of 5 (which I verified).

Do the same sort of calculations for 7⁷, 9⁷, and 11⁷, and see what you get.

 

[zgozvrm]

...what did I do wrong?

You can't add bases like that, for instance, [itex]3^2 + 7^2 = 9 + 49 = 58[/tex]<br /> (NOT [itex]3^2 + 7^2 = (3 + 7)^2 = 10^2 = 100[/tex])[/itex][/itex]

 

[morrowcosom]

Let's take these powers in small steps, starting with 57.
5 5 (mod 24)
52 = 25 1 (mod 24)

So 57 = 5 * 56 = 5 * (52)3 5 * 13 (mod 24) = 5 (mod 24).

If you divide 57 by 24, you get a remainder of 5 (which I verified).

Do the same sort of calculations for 77, 97, and 117, and see what you get.

Original problem: Calculate (5^7)-(7^7)+(9^7)-(11^7) in Mod24

I started out with 5 that you figured out and worked the other numbers the same way:

-7^7= -7*7^6= -7(-7^2)^3= -7*1^3= -7 (mod 24)

9^7= 9*9^6= 9(9^2)^3= 9*9^3= 81= 9 (mod 24)

11^7= 11*11^6= 11(11^2)^3= 11*1^3= 11 (mod 24)

So, I ended up with 5-7+9+11=18, which is still the wrong answer according to the program.
Where did I mess up?

 

[Mark44]

You have a sign error: you should have 5 - 7 + 9 - 11