Question 1205778
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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)
*[tex \Large a^{p} \equiv a (\text{mod } p)] 
'a' is an integer
p is prime


Special case of FLT
*[tex \Large a^{p-1} \equiv 1 (\text{mod } p)]
where p is <u>not</u> a factor of 'a' 



*[tex \Large 10^{12} \equiv 1 (\text{mod } 13)] Due to the special case of FLT


*[tex \Large \left(10^{12}\right)^{20} \equiv 1^{20} (\text{mod } 13)] Raise both sides to the 20th power


*[tex \Large 10^{12*20} \equiv 1 (\text{mod } 13)] Use the rule that (a^b)^c = a^(b*c)

 
*[tex \Large 10^{240} \equiv 1 (\text{mod } 13)] 


*[tex \Large 10*10^{240} \equiv 10*1 (\text{mod } 13)] Multiply both sides by 10


*[tex \Large 10^1*10^{240} \equiv 10 (\text{mod } 13)] 


*[tex \Large 10^{1+240} \equiv 10 (\text{mod } 13)] Use the rule that a^b*a^c = a^(b+c)


*[tex \Large 10^{241} \equiv 10 (\text{mod } 13)] 



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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. 


*[tex \Large 10^{1} \equiv 10 (\text{mod } 13)]


*[tex \Large 10^{2} \equiv 100 \equiv 9  (\text{mod } 13)] because 100/13 = 7 remainder 9. We only care about the remainder.


*[tex \Large 10^{3} \equiv 10*10^2 \equiv 10*9 \equiv 90 \equiv 12  (\text{mod } 13)] because 90/13 = 6 remainder 12. 


*[tex \Large 10^{4} \equiv 10*10^3 \equiv 10*12 \equiv 120 \equiv 3  (\text{mod } 13)] because 120/13 = 9 remainder 3


You can keep this process going. 
I'll use a spreadsheet to extend this list. I'm using a command called <font color=red>Mod</font>
Example: Type in <font color=red>=Mod(100,13)</font> to get 9
<table border = "1" cellpadding = "5"><tr><td>n</td><td>10^n mod 13</td></tr><tr><td>1</td><td>10</td></tr><tr><td>2</td><td>9</td></tr><tr><td>3</td><td>12</td></tr><tr><td>4</td><td>3</td></tr><tr><td>5</td><td>4</td></tr><tr><td>6</td><td>1</td></tr><tr><td>7</td><td>10</td></tr><tr><td>8</td><td>9</td></tr><tr><td>9</td><td>12</td></tr><tr><td>10</td><td>3</td></tr></table>
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 
*[tex \Large 10^{241} \equiv 10  (\text{mod } 13)]


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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


*[tex \Large 10^{2} \equiv 9 (\text{mod } 13)] mentioned in the previous section


*[tex \Large  10^{4} \equiv \left(10^{2}\right)^2 \equiv 3 (\text{mod } 13)] mentioned in the previous section


*[tex \Large  10^{8} \equiv \left(10^{4}\right)^2 \equiv \left(3\right)^2   \equiv 9 (\text{mod } 13)]


*[tex \Large  10^{16} \equiv \left(10^{8}\right)^2 \equiv \left(9\right)^2   \equiv 81  \equiv 3 (\text{mod } 13)] because 81/13 = 6 remainder 3.


Keep this process going to generate this table
<table border = "1" cellpadding = "5"><tr><td>n</td><td>2^n</td><td>10^(2^n) mod 13</td></tr><tr><td>1</td><td>2</td><td>9</td></tr><tr><td>2</td><td>4</td><td>3</td></tr><tr><td>3</td><td>8</td><td>9</td></tr><tr><td>4</td><td>16</td><td>3</td></tr><tr><td>5</td><td>32</td><td>9</td></tr><tr><td>6</td><td>64</td><td>3</td></tr><tr><td>7</td><td>128</td><td>9</td></tr><tr><td>8</td>
<td>256</td><td>3</td></tr></table>
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
*[tex \Large 10^{241} \equiv 10^{128+64+32+16+1} (\text{mod } 13)]


*[tex \Large 10^{241} \equiv 10^{128}*10^{64}*10^{32}*10^{16}*10^{1} (\text{mod } 13)]


*[tex \Large 10^{241} \equiv 9*3*9*3*10 (\text{mod } 13)] Use the table shown above. For example, 10^128 = 9 (mod 13)


*[tex \Large 10^{241} \equiv (9*3)*(9*3)*10 (\text{mod } 13)]


*[tex \Large 10^{241} \equiv (27)*(27)*10 (\text{mod } 13)]


*[tex \Large 10^{241} \equiv (1)*(1)*10 (\text{mod } 13)] because 27/13 = 2 remainder 1


*[tex \Large 10^{241} \equiv 10 (\text{mod } 13)]


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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
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