You have the correct line of thinking and correct answer. Nice work.
This first section basically restates what you mentioned.
The next two sections will go over other approaches.
Fermat's Little Theorem (FLT)
'a' is an integer
p is prime
Special case of FLT
where p is not a factor of 'a'
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A 2nd approach:
Let's look at powers of 10^n (mod 13) to see if we can spot any patterns.
You can keep this process going.
I'll use a spreadsheet to extend this list. I'm using a command called Mod
Example: Type in =Mod(100,13) to get 9
| n | 10^n mod 13 |
| 1 | 10 |
| 2 | 9 |
| 3 | 12 |
| 4 | 3 |
| 5 | 4 |
| 6 | 1 |
| 7 | 10 |
| 8 | 9 |
| 9 | 12 |
| 10 | 3 |
I'll leave the scratch work for the student to do.
The pattern is 10, 9, 12, 3, 4, 1, 10, 9...
Once reaching remainder 1, the pattern repeats over again.
This is simply because 1 times any number is that number.
There are 6 items in the set {10, 9, 12, 3, 4, 1}
So this means we'll divide 241 over 6 to look at the remainder (yes it seems a bit strange to change the modulus all of a sudden)
241/6 = 40 remainder 1
The remainder 1 means we'll look at the 1st slot of that 6 number repeating pattern. That 1st slot is 10.
Therefore we found another way to determine that
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A 3rd approach:
Let's look at terms of the form 10^(2^n) where n is a positive integer
Effectively we're repeatedly squaring powers of 10, since something like 10^4 is the squaring of 10^2; or 10^8 is the squaring of 10^4, etc
Keep this process going to generate this table
| n | 2^n | 10^(2^n) mod 13 |
| 1 | 2 | 9 |
| 2 | 4 | 3 |
| 3 | 8 | 9 |
| 4 | 16 | 3 |
| 5 | 32 | 9 |
| 6 | 64 | 3 |
| 7 | 128 | 9 |
| 8 | 256 | 3 |
We have a much nicer pattern going on here.
The remainder is either 9 or 3 depending if n is odd or even in that order.
Technically a table isn't needed, but it could be nice to have.
This is so we can break 241 into its binary components so to speak.
Refer to tutorials on how to convert from base 10 to base 2.
241 base 10 = 11110001 base 2
241 base 10 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0
241 base 10 = 1*128 + 1*64 + 1*32 + 1*16 + 0*8 + 0*4 + 0*2 + 1*1
241 base 10 = 128 + 64 + 32 + 16 + 1
241 = 128 + 64 + 32 + 16 + 1
Then notice the following computation
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FLT is perhaps the best choice since it appears your professor has introduced it and is likely expecting you to use it; however, it's still good practice to be able to tackle problems in various ways.
For more practice, check out this similar question
https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1205936.html