Modular Arithmetic: Solve (21999+31998+51997) Divided by 7

[triac]

Homework Statement

Okay, so I'm going to find the smallest positive remainder of (2¹⁹⁹⁹+3¹⁹⁹⁸+5¹⁹⁹⁷) divided by seven.

Homework Equations

The Attempt at a Solution

Well, I did like this:
2³ is congruent to 1 (mod 7). Therefore, 2¹⁹⁹⁹= (2³)^1999/3 is congruent to 1 (mod 7).
3³ is congruent to (-1) (mod 7). Therefore, 3¹⁹⁹⁸=(3³)⁶⁶⁶ is congruent to (-1)⁶⁶⁶=1 (mod 7).
5³ is congruent to (-1) (mod 7). Therefore, 5¹⁹⁹⁷=(5³)⁶⁶⁵*25 is congruent to (-1)⁶⁶⁵*4 = -4 (mod 7)

So, all in all the remainder should be two. However it says in the key that it is six, and I can't see where I'm wrong. Got any suggestions?

 

[Mark44]

Homework Statement

Okay, so I'm going to find the smallest positive remainder of (2¹⁹⁹⁹+3¹⁹⁹⁸+5¹⁹⁹⁷) divided by seven.

Homework Equations

The Attempt at a Solution

Well, I did like this:
2³ is congruent to 1 (mod 7). Therefore, 2¹⁹⁹⁹= (2³)^1999/3 is congruent to 1 (mod 7).
3³ is congruent to (-1) (mod 7). Therefore, 3¹⁹⁹⁸=(3³)⁶⁶⁶ is congruent to (-1)⁶⁶⁶=1 (mod 7).
5³ is congruent to (-1) (mod 7). Therefore, 5¹⁹⁹⁷=(5³)⁶⁶⁵*25 is congruent to (-1)⁶⁶⁵*4 = -4 (mod 7)

So, all in all the remainder should be two. However it says in the key that it is six, and I can't see where I'm wrong. Got any suggestions?

Here's what I get:
2³ is congruent to 1 (mod 7). Therefore, 2¹⁹⁹⁹= (2³)⁶⁶⁶*2 is congruent to 2 (mod 7) (not 1 mod 7 as you had).
3³ is congruent to (-1) (mod 7). Therefore, 3¹⁹⁹⁸=(3³)⁶⁶⁶ is congruent to (-1)⁶⁶⁶=1 (mod 7).
5³ is congruent to (-1) (mod 7). Therefore, 5¹⁹⁹⁷=(5³)⁶⁶⁵*25 is congruent to (-1)⁶⁶⁵*4 = -4 (mod 7) = 3 mod 7.

Add 'em up and you get 6 mod 7.

 

[triac]

Ok, thanks a lot!
I just wonder, why is it wrong to do the way I did, why doesn't it work?

 

[tiny-tim]

Hi triac!

I just wonder, why is it wrong to do the way I did, why doesn't it work?

Because 1999/3 isn't a whole number.