Evaluation

Input	Output
Find the least odd prime factor of $2019^8+1$ .	Qwen: QwQ 32B Preview PublicPass So I need to find the least odd prime factor of ( 2019^8 + 1 ). Hmm, that seems pretty big, but I think I can break it down step by step. First, I know that finding prime factors of large exponents can be tricky, but there might be some properties or theorems that can help simplify the process. Since the expression is ( 2019^8 + 1 ), maybe I can look for patterns or ways to factor this expression. I recall that expressions of the form ( a^n + b^n ) can sometimes be factored, especially when ( n ) is a power of 2. In this case, ( n = 8 ), which is ( 2^3 ), so that might be useful. Let me see if I can factor ( 2019^8 + 1 ). I know that ( a^4 + 1 ) can be factored as ( (a^4 + 1) = (a^2 + \sqrt{2}a + 1)(a^2 - \sqrt{2}a + 1) ), but that involves irrational numbers, which isn't helpful for factoring into integers. Maybe there's another way. Wait, perhaps I can write ( 2019^8 + 1 ) as ( (2019^4)^2 + 1^2 ), which resembles the sum of squares. I know that over the integers, sums of squares don't factor nicely, but maybe there's a different approach. Another thought: maybe I can consider smaller exponents first and see if there's a pattern or a recurring factor that appears. Let's try computing ( 2019^2 + 1 ), ( 2019^4 + 1 ), and so on, and see if I can find any common factors or patterns. First, calculate ( 2019^2 + 1 ): ( 2019^2 = 2019 \times 2019 ). I don't need the exact value right now; maybe I can just keep it as ( 2019^2 ) for now. So, ( 2019^2 + 1 = 2019^2 + 1 ). Doesn't seem immediately helpful. Next, ( 2019^4 + 1 ). Again, it's similar to the previous expression, and I don't see an obvious factorization. Maybe I should try to see if there are any small prime numbers that divide ( 2019^8 + 1 ). Since I'm looking for the least odd prime factor, I can start testing small primes like 3, 5, 7, 11, and so on, to see if any of them divide ( 2019^8 + 1 ). Let's start with 3. To check if 3 divides ( 2019^8 + 1 ), I can compute ( 2019^8 + 1 ) modulo 3. First, find ( 2019 ) modulo 3. Since ( 2019 = 3 \times 673 ), ( 2019 \equiv 0 \pmod{3} ). Therefore, ( 2019^8 \equiv 0^8 \equiv 0 \pmod{3} ), and ( 2019^8 + 1 \equiv 0 + 1 \equiv 1 \pmod{3} ). So, 3 does not divide ( 2019^8 + 1 ). Next, check 5. Compute ( 2019 ) modulo 5. ( 2019 \div 5 = 403 ) with a remainder of 4, so ( 2019 \equiv 4 \pmod{5} ). Then, ( 2019^8 \equiv 4^8 \pmod{5} ). Since ( 4 \equiv -1 \pmod{5} ), ( 4^8 \equiv (-1)^8 \equiv 1 \pmod{5} ). Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{5} ), so 5 does not divide ( 2019^8 + 1 ). Now, check 7. Find ( 2019 ) modulo 7. ( 2019 \div 7 = 288 ) with a remainder of 3, so ( 2019 \equiv 3 \pmod{7} ). Then, ( 2019^8 \equiv 3^8 \pmod{7} ). Let's compute ( 3^8 ) modulo 7. First, find the pattern of powers of 3 modulo 7: ( 3^1 \equiv 3 \pmod{7} ) ( 3^2 \equiv 9 \equiv 2 \pmod{7} ) ( 3^3 \equiv 6 \pmod{7} ) ( 3^4 \equiv 18 \equiv 4 \pmod{7} ) ( 3^5 \equiv 12 \equiv 5 \pmod{7} ) ( 3^6 \equiv 15 \equiv 1 \pmod{7} ) Since ( 3^6 \equiv 1 \pmod{7} ), then ( 3^8 = 3^{6+2} = 3^6 \times 3^2 \equiv 1 \times 2 \equiv 2 \pmod{7} ). Therefore, ( 2019^8 + 1 \equiv 2 + 1 \equiv 3 \pmod{7} ), so 7 does not divide ( 2019^8 + 1 ). Next, check 11. Compute ( 2019 ) modulo 11. ( 2019 \div 11 = 183 ) with a remainder of 6, so ( 2019 \equiv 6 \pmod{11} ). Then, ( 2019^8 \equiv 6^8 \pmod{11} ). Let's find a pattern for powers of 6 modulo 11: ( 6^1 \equiv 6 \pmod{11} ) ( 6^2 \equiv 36 \equiv 3 \pmod{11} ) ( 6^3 \equiv 18 \equiv 7 \pmod{11} ) ( 6^4 \equiv 42 \equiv 9 \pmod{11} ) ( 6^5 \equiv 54 \equiv 10 \pmod{11} ) ( 6^6 \equiv 60 \equiv 5 \pmod{11} ) ( 6^7 \equiv 30 \equiv 8 \pmod{11} ) ( 6^8 \equiv 48 \equiv 4 \pmod{11} ) Therefore, ( 2019^8 + 1 \equiv 4 + 1 \equiv 5 \pmod{11} ), so 11 does not divide ( 2019^8 + 1 ). Continuing this process seems tedious. Maybe there's a better way to approach this problem. Let me think differently. Perhaps I can consider the expression ( 2019^8 + 1 ) and see if it can be expressed in a form that reveals its prime factors. I recall that for expressions like ( a^4 + 4b^4 ), there's a Sophie Germain identity: ( a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) ). Maybe I can manipulate ( 2019^8 + 1 ) into a similar form. Wait, ( 2019^8 + 1 = (2019^4)^2 + 1^2 ). This resembles the sum of squares, but over integers, sums of squares don't factor nicely. Maybe I can consider complex numbers or some other algebraic identities, but that might not help in finding integer prime factors. Alternatively, perhaps I can consider modulo properties to find possible prime factors. For example, if ( p ) is an odd prime that divides ( 2019^8 + 1 ), then ( 2019^8 \equiv -1 \pmod{p} ), which implies ( 2019^{16} \equiv 1 \pmod{p} ). This means that the order of 2019 modulo ( p ) divides 16, and since ( 2019^8 \equiv -1 \pmod{p} ), the order can't divide 8. Therefore, the order must be 16. By Fermat's Little Theorem, the order of 2019 modulo ( p ) divides ( p - 1 ). Therefore, ( p - 1 ) must be divisible by 16. So, ( p \equiv 1 \pmod{16} ). This narrows down the possible primes ( p ) that could divide ( 2019^8 + 1 ) to those primes congruent to 1 modulo 16. Let me list some primes that are 1 modulo 16: 17, 97, 113, 193, 241, 257, etc. Now, I can test these primes to see if any of them divide ( 2019^8 + 1 ). Start with 17. Compute ( 2019 ) modulo 17. First, find ( 2019 \div 17 ). 17 times 118 is 2006, so ( 2019 - 2006 = 13 ). Therefore, ( 2019 \equiv 13 \pmod{17} ). Then, ( 2019^8 \equiv 13^8 \pmod{17} ). Compute ( 13^2 \equiv 169 \equiv 16 \pmod{17} ) ( 13^4 = (13^2)^2 \equiv 16^2 \equiv 256 \equiv 1 \pmod{17} ) ( 13^8 = (13^4)^2 \equiv 1^2 \equiv 1 \pmod{17} ) Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{17} ), so 17 does not divide ( 2019^8 + 1 ). Next, check 97. Compute ( 2019 ) modulo 97. First, find how many times 97 fits into 2019. 97 times 20 is 1940, and 2019 - 1940 = 79. So, ( 2019 \equiv 79 \pmod{97} ). Then, ( 2019^8 \equiv 79^8 \pmod{97} ). This seems a bit cumbersome to compute directly. Maybe I can find a pattern or use Fermat's Little Theorem, which states that ( a^{p-1} \equiv 1 \pmod{p} ) if ( p ) is prime and ( p ) does not divide ( a ). Here, ( p = 97 ), so ( 79^{96} \equiv 1 \pmod{97} ). But I need ( 79^8 ), which is not directly helpful. Maybe I can compute ( 79^2 ), then ( 79^4 ), and so on. First, compute ( 79^2 \pmod{97} ): ( 79^2 = 6241 ). Now, divide 6241 by 97. 97 times 64 is 6208, so 6241 - 6208 = 33. Therefore, ( 79^2 \equiv 33 \pmod{97} ). Next, ( 79^4 = (79^2)^2 \equiv 33^2 \pmod{97} ). ( 33^2 = 1089 ). 97 times 11 is 1067, so 1089 - 1067 = 22. Therefore, ( 79^4 \equiv 22 \pmod{97} ). Then, ( 79^8 = (79^4)^2 \equiv 22^2 \pmod{97} ). ( 22^2 = 484 ). 97 times 4 is 388, so 484 - 388 = 96. Therefore, ( 79^8 \equiv 96 \pmod{97} ). Thus, ( 2019^8 + 1 \equiv 96 + 1 \equiv 97 \equiv 0 \pmod{97} ). So, 97 divides ( 2019^8 + 1 ). Since 97 is the smallest prime congruent to 1 modulo 16 that we've checked and it divides ( 2019^8 + 1 ), it must be the least odd prime factor of ( 2019^8 + 1 ). Therefore, the least odd prime factor is 97. Final Answer [ \boxed{97} ]

Input

Output

Find the least odd prime factor of $2019^8+1$ .

PublicPass

So I need to find the least odd prime factor of ( 2019^8 + 1 ). Hmm, that seems pretty big, but I think I can break it down step by step.

First, I know that finding prime factors of large exponents can be tricky, but there might be some properties or theorems that can help simplify the process. Since the expression is ( 2019^8 + 1 ), maybe I can look for patterns or ways to factor this expression.

I recall that expressions of the form ( a^n + b^n ) can sometimes be factored, especially when ( n ) is a power of 2. In this case, ( n = 8 ), which is ( 2^3 ), so that might be useful.

Let me see if I can factor ( 2019^8 + 1 ). I know that ( a^4 + 1 ) can be factored as ( (a^4 + 1) = (a^2 + \sqrt{2}a + 1)(a^2 - \sqrt{2}a + 1) ), but that involves irrational numbers, which isn't helpful for factoring into integers. Maybe there's another way.

Wait, perhaps I can write ( 2019^8 + 1 ) as ( (2019^4)^2 + 1^2 ), which resembles the sum of squares. I know that over the integers, sums of squares don't factor nicely, but maybe there's a different approach.

Another thought: maybe I can consider smaller exponents first and see if there's a pattern or a recurring factor that appears.

Let's try computing ( 2019^2 + 1 ), ( 2019^4 + 1 ), and so on, and see if I can find any common factors or patterns.

First, calculate ( 2019^2 + 1 ):

( 2019^2 = 2019 \times 2019 ). I don't need the exact value right now; maybe I can just keep it as ( 2019^2 ) for now.

So, ( 2019^2 + 1 = 2019^2 + 1 ). Doesn't seem immediately helpful.

Next, ( 2019^4 + 1 ). Again, it's similar to the previous expression, and I don't see an obvious factorization.

Maybe I should try to see if there are any small prime numbers that divide ( 2019^8 + 1 ). Since I'm looking for the least odd prime factor, I can start testing small primes like 3, 5, 7, 11, and so on, to see if any of them divide ( 2019^8 + 1 ).

Let's start with 3. To check if 3 divides ( 2019^8 + 1 ), I can compute ( 2019^8 + 1 ) modulo 3.

First, find ( 2019 ) modulo 3. Since ( 2019 = 3 \times 673 ), ( 2019 \equiv 0 \pmod{3} ). Therefore, ( 2019^8 \equiv 0^8 \equiv 0 \pmod{3} ), and ( 2019^8 + 1 \equiv 0 + 1 \equiv 1 \pmod{3} ). So, 3 does not divide ( 2019^8 + 1 ).

Next, check 5. Compute ( 2019 ) modulo 5. ( 2019 \div 5 = 403 ) with a remainder of 4, so ( 2019 \equiv 4 \pmod{5} ). Then, ( 2019^8 \equiv 4^8 \pmod{5} ). Since ( 4 \equiv -1 \pmod{5} ), ( 4^8 \equiv (-1)^8 \equiv 1 \pmod{5} ). Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{5} ), so 5 does not divide ( 2019^8 + 1 ).

Now, check 7. Find ( 2019 ) modulo 7. ( 2019 \div 7 = 288 ) with a remainder of 3, so ( 2019 \equiv 3 \pmod{7} ). Then, ( 2019^8 \equiv 3^8 \pmod{7} ). Let's compute ( 3^8 ) modulo 7.

First, find the pattern of powers of 3 modulo 7:

( 3^1 \equiv 3 \pmod{7} )

( 3^2 \equiv 9 \equiv 2 \pmod{7} )

( 3^3 \equiv 6 \pmod{7} )

( 3^4 \equiv 18 \equiv 4 \pmod{7} )

( 3^5 \equiv 12 \equiv 5 \pmod{7} )

( 3^6 \equiv 15 \equiv 1 \pmod{7} )

Since ( 3^6 \equiv 1 \pmod{7} ), then ( 3^8 = 3^{6+2} = 3^6 \times 3^2 \equiv 1 \times 2 \equiv 2 \pmod{7} ). Therefore, ( 2019^8 + 1 \equiv 2 + 1 \equiv 3 \pmod{7} ), so 7 does not divide ( 2019^8 + 1 ).

Next, check 11. Compute ( 2019 ) modulo 11. ( 2019 \div 11 = 183 ) with a remainder of 6, so ( 2019 \equiv 6 \pmod{11} ). Then, ( 2019^8 \equiv 6^8 \pmod{11} ). Let's find a pattern for powers of 6 modulo 11:

( 6^1 \equiv 6 \pmod{11} )

( 6^2 \equiv 36 \equiv 3 \pmod{11} )

( 6^3 \equiv 18 \equiv 7 \pmod{11} )

( 6^4 \equiv 42 \equiv 9 \pmod{11} )

( 6^5 \equiv 54 \equiv 10 \pmod{11} )

( 6^6 \equiv 60 \equiv 5 \pmod{11} )

( 6^7 \equiv 30 \equiv 8 \pmod{11} )

( 6^8 \equiv 48 \equiv 4 \pmod{11} )

Therefore, ( 2019^8 + 1 \equiv 4 + 1 \equiv 5 \pmod{11} ), so 11 does not divide ( 2019^8 + 1 ).

Continuing this process seems tedious. Maybe there's a better way to approach this problem.

Let me think differently. Perhaps I can consider the expression ( 2019^8 + 1 ) and see if it can be expressed in a form that reveals its prime factors.

I recall that for expressions like ( a^4 + 4b^4 ), there's a Sophie Germain identity: ( a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) ). Maybe I can manipulate ( 2019^8 + 1 ) into a similar form.

Wait, ( 2019^8 + 1 = (2019^4)^2 + 1^2 ). This resembles the sum of squares, but over integers, sums of squares don't factor nicely. Maybe I can consider complex numbers or some other algebraic identities, but that might not help in finding integer prime factors.

Alternatively, perhaps I can consider modulo properties to find possible prime factors. For example, if ( p ) is an odd prime that divides ( 2019^8 + 1 ), then ( 2019^8 \equiv -1 \pmod{p} ), which implies ( 2019^{16} \equiv 1 \pmod{p} ). This means that the order of 2019 modulo ( p ) divides 16, and since ( 2019^8 \equiv -1 \pmod{p} ), the order can't divide 8. Therefore, the order must be 16.

By Fermat's Little Theorem, the order of 2019 modulo ( p ) divides ( p - 1 ). Therefore, ( p - 1 ) must be divisible by 16. So, ( p \equiv 1 \pmod{16} ).

This narrows down the possible primes ( p ) that could divide ( 2019^8 + 1 ) to those primes congruent to 1 modulo 16.

Let me list some primes that are 1 modulo 16: 17, 97, 113, 193, 241, 257, etc.

Now, I can test these primes to see if any of them divide ( 2019^8 + 1 ).

Start with 17. Compute ( 2019 ) modulo 17.

First, find ( 2019 \div 17 ). 17 times 118 is 2006, so ( 2019 - 2006 = 13 ). Therefore, ( 2019 \equiv 13 \pmod{17} ).

Then, ( 2019^8 \equiv 13^8 \pmod{17} ).

Compute ( 13^2 \equiv 169 \equiv 16 \pmod{17} )

( 13^4 = (13^2)^2 \equiv 16^2 \equiv 256 \equiv 1 \pmod{17} )

( 13^8 = (13^4)^2 \equiv 1^2 \equiv 1 \pmod{17} )

Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{17} ), so 17 does not divide ( 2019^8 + 1 ).

Next, check 97. Compute ( 2019 ) modulo 97.

First, find how many times 97 fits into 2019. 97 times 20 is 1940, and 2019 - 1940 = 79. So, ( 2019 \equiv 79 \pmod{97} ).

Then, ( 2019^8 \equiv 79^8 \pmod{97} ).

This seems a bit cumbersome to compute directly. Maybe I can find a pattern or use Fermat's Little Theorem, which states that ( a^{p-1} \equiv 1 \pmod{p} ) if ( p ) is prime and ( p ) does not divide ( a ).

Here, ( p = 97 ), so ( 79^{96} \equiv 1 \pmod{97} ). But I need ( 79^8 ), which is not directly helpful. Maybe I can compute ( 79^2 ), then ( 79^4 ), and so on.

First, compute ( 79^2 \pmod{97} ):

( 79^2 = 6241 ). Now, divide 6241 by 97.

97 times 64 is 6208, so 6241 - 6208 = 33. Therefore, ( 79^2 \equiv 33 \pmod{97} ).

Next, ( 79^4 = (79^2)^2 \equiv 33^2 \pmod{97} ).

( 33^2 = 1089 ). 97 times 11 is 1067, so 1089 - 1067 = 22. Therefore, ( 79^4 \equiv 22 \pmod{97} ).

Then, ( 79^8 = (79^4)^2 \equiv 22^2 \pmod{97} ).

( 22^2 = 484 ). 97 times 4 is 388, so 484 - 388 = 96. Therefore, ( 79^8 \equiv 96 \pmod{97} ).

Thus, ( 2019^8 + 1 \equiv 96 + 1 \equiv 97 \equiv 0 \pmod{97} ).

So, 97 divides ( 2019^8 + 1 ). Since 97 is the smallest prime congruent to 1 modulo 16 that we've checked and it divides ( 2019^8 + 1 ), it must be the least odd prime factor of ( 2019^8 + 1 ).

Therefore, the least odd prime factor is 97.

Final Answer

[ \boxed{97} ]

Evaluation ID: 2932d8ae-d731-4a2d-b1f5-3c445ce60cd3
Created: November 28, 2024

Evaluation Rating

Binary Pass/Fail: Pass

Additional Feedback: The student's submitted answer is correct, detailed, and includes a comprehensive exploration and calculation leading to the correct answer of 97, which matches the expert answer.

Raw Output

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        "content": "So I need to find the least odd prime factor of \\( 2019^8 + 1 \\). Hmm, that seems pretty big, but I think I can break it down step by step.\n\nFirst, I know that finding prime factors of large exponents can be tricky, but there might be some properties or theorems that can help simplify the process. Since the expression is \\( 2019^8 + 1 \\), maybe I can look for patterns or ways to factor this expression.\n\nI recall that expressions of the form \\( a^n + b^n \\) can sometimes be factored, especially when \\( n \\) is a power of 2. In this case, \\( n = 8 \\), which is \\( 2^3 \\), so that might be useful.\n\nLet me see if I can factor \\( 2019^8 + 1 \\). I know that \\( a^4 + 1 \\) can be factored as \\( (a^4 + 1) = (a^2 + \\sqrt{2}a + 1)(a^2 - \\sqrt{2}a + 1) \\), but that involves irrational numbers, which isn't helpful for factoring into integers. Maybe there's another way.\n\nWait, perhaps I can write \\( 2019^8 + 1 \\) as \\( (2019^4)^2 + 1^2 \\), which resembles the sum of squares. I know that over the integers, sums of squares don't factor nicely, but maybe there's a different approach.\n\nAnother thought: maybe I can consider smaller exponents first and see if there's a pattern or a recurring factor that appears.\n\nLet's try computing \\( 2019^2 + 1 \\), \\( 2019^4 + 1 \\), and so on, and see if I can find any common factors or patterns.\n\nFirst, calculate \\( 2019^2 + 1 \\):\n\n\\( 2019^2 = 2019 \\times 2019 \\). I don't need the exact value right now; maybe I can just keep it as \\( 2019^2 \\) for now.\n\nSo, \\( 2019^2 + 1 = 2019^2 + 1 \\). Doesn't seem immediately helpful.\n\nNext, \\( 2019^4 + 1 \\). Again, it's similar to the previous expression, and I don't see an obvious factorization.\n\nMaybe I should try to see if there are any small prime numbers that divide \\( 2019^8 + 1 \\). Since I'm looking for the least odd prime factor, I can start testing small primes like 3, 5, 7, 11, and so on, to see if any of them divide \\( 2019^8 + 1 \\).\n\nLet's start with 3. To check if 3 divides \\( 2019^8 + 1 \\), I can compute \\( 2019^8 + 1 \\) modulo 3.\n\nFirst, find \\( 2019 \\) modulo 3. Since \\( 2019 = 3 \\times 673 \\), \\( 2019 \\equiv 0 \\pmod{3} \\). Therefore, \\( 2019^8 \\equiv 0^8 \\equiv 0 \\pmod{3} \\), and \\( 2019^8 + 1 \\equiv 0 + 1 \\equiv 1 \\pmod{3} \\). So, 3 does not divide \\( 2019^8 + 1 \\).\n\nNext, check 5. Compute \\( 2019 \\) modulo 5. \\( 2019 \\div 5 = 403 \\) with a remainder of 4, so \\( 2019 \\equiv 4 \\pmod{5} \\). Then, \\( 2019^8 \\equiv 4^8 \\pmod{5} \\). Since \\( 4 \\equiv -1 \\pmod{5} \\), \\( 4^8 \\equiv (-1)^8 \\equiv 1 \\pmod{5} \\). Therefore, \\( 2019^8 + 1 \\equiv 1 + 1 \\equiv 2 \\pmod{5} \\), so 5 does not divide \\( 2019^8 + 1 \\).\n\nNow, check 7. Find \\( 2019 \\) modulo 7. \\( 2019 \\div 7 = 288 \\) with a remainder of 3, so \\( 2019 \\equiv 3 \\pmod{7} \\). Then, \\( 2019^8 \\equiv 3^8 \\pmod{7} \\). Let's compute \\( 3^8 \\) modulo 7.\n\nFirst, find the pattern of powers of 3 modulo 7:\n\n\\( 3^1 \\equiv 3 \\pmod{7} \\)\n\n\\( 3^2 \\equiv 9 \\equiv 2 \\pmod{7} \\)\n\n\\( 3^3 \\equiv 6 \\pmod{7} \\)\n\n\\( 3^4 \\equiv 18 \\equiv 4 \\pmod{7} \\)\n\n\\( 3^5 \\equiv 12 \\equiv 5 \\pmod{7} \\)\n\n\\( 3^6 \\equiv 15 \\equiv 1 \\pmod{7} \\)\n\nSince \\( 3^6 \\equiv 1 \\pmod{7} \\), then \\( 3^8 = 3^{6+2} = 3^6 \\times 3^2 \\equiv 1 \\times 2 \\equiv 2 \\pmod{7} \\). Therefore, \\( 2019^8 + 1 \\equiv 2 + 1 \\equiv 3 \\pmod{7} \\), so 7 does not divide \\( 2019^8 + 1 \\).\n\nNext, check 11. Compute \\( 2019 \\) modulo 11. \\( 2019 \\div 11 = 183 \\) with a remainder of 6, so \\( 2019 \\equiv 6 \\pmod{11} \\). Then, \\( 2019^8 \\equiv 6^8 \\pmod{11} \\). Let's find a pattern for powers of 6 modulo 11:\n\n\\( 6^1 \\equiv 6 \\pmod{11} \\)\n\n\\( 6^2 \\equiv 36 \\equiv 3 \\pmod{11} \\)\n\n\\( 6^3 \\equiv 18 \\equiv 7 \\pmod{11} \\)\n\n\\( 6^4 \\equiv 42 \\equiv 9 \\pmod{11} \\)\n\n\\( 6^5 \\equiv 54 \\equiv 10 \\pmod{11} \\)\n\n\\( 6^6 \\equiv 60 \\equiv 5 \\pmod{11} \\)\n\n\\( 6^7 \\equiv 30 \\equiv 8 \\pmod{11} \\)\n\n\\( 6^8 \\equiv 48 \\equiv 4 \\pmod{11} \\)\n\nTherefore, \\( 2019^8 + 1 \\equiv 4 + 1 \\equiv 5 \\pmod{11} \\), so 11 does not divide \\( 2019^8 + 1 \\).\n\nContinuing this process seems tedious. Maybe there's a better way to approach this problem.\n\nLet me think differently. Perhaps I can consider the expression \\( 2019^8 + 1 \\) and see if it can be expressed in a form that reveals its prime factors.\n\nI recall that for expressions like \\( a^4 + 4b^4 \\), there's a Sophie Germain identity: \\( a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) \\). Maybe I can manipulate \\( 2019^8 + 1 \\) into a similar form.\n\nWait, \\( 2019^8 + 1 = (2019^4)^2 + 1^2 \\). This resembles the sum of squares, but over integers, sums of squares don't factor nicely. Maybe I can consider complex numbers or some other algebraic identities, but that might not help in finding integer prime factors.\n\nAlternatively, perhaps I can consider modulo properties to find possible prime factors. For example, if \\( p \\) is an odd prime that divides \\( 2019^8 + 1 \\), then \\( 2019^8 \\equiv -1 \\pmod{p} \\), which implies \\( 2019^{16} \\equiv 1 \\pmod{p} \\). This means that the order of 2019 modulo \\( p \\) divides 16, and since \\( 2019^8 \\equiv -1 \\pmod{p} \\), the order can't divide 8. Therefore, the order must be 16.\n\nBy Fermat's Little Theorem, the order of 2019 modulo \\( p \\) divides \\( p - 1 \\). Therefore, \\( p - 1 \\) must be divisible by 16. So, \\( p \\equiv 1 \\pmod{16} \\).\n\nThis narrows down the possible primes \\( p \\) that could divide \\( 2019^8 + 1 \\) to those primes congruent to 1 modulo 16.\n\nLet me list some primes that are 1 modulo 16: 17, 97, 113, 193, 241, 257, etc.\n\nNow, I can test these primes to see if any of them divide \\( 2019^8 + 1 \\).\n\nStart with 17. Compute \\( 2019 \\) modulo 17.\n\nFirst, find \\( 2019 \\div 17 \\). 17 times 118 is 2006, so \\( 2019 - 2006 = 13 \\). Therefore, \\( 2019 \\equiv 13 \\pmod{17} \\).\n\nThen, \\( 2019^8 \\equiv 13^8 \\pmod{17} \\).\n\nCompute \\( 13^2 \\equiv 169 \\equiv 16 \\pmod{17} \\)\n\n\\( 13^4 = (13^2)^2 \\equiv 16^2 \\equiv 256 \\equiv 1 \\pmod{17} \\)\n\n\\( 13^8 = (13^4)^2 \\equiv 1^2 \\equiv 1 \\pmod{17} \\)\n\nTherefore, \\( 2019^8 + 1 \\equiv 1 + 1 \\equiv 2 \\pmod{17} \\), so 17 does not divide \\( 2019^8 + 1 \\).\n\nNext, check 97. Compute \\( 2019 \\) modulo 97.\n\nFirst, find how many times 97 fits into 2019. 97 times 20 is 1940, and 2019 - 1940 = 79. So, \\( 2019 \\equiv 79 \\pmod{97} \\).\n\nThen, \\( 2019^8 \\equiv 79^8 \\pmod{97} \\).\n\nThis seems a bit cumbersome to compute directly. Maybe I can find a pattern or use Fermat's Little Theorem, which states that \\( a^{p-1} \\equiv 1 \\pmod{p} \\) if \\( p \\) is prime and \\( p \\) does not divide \\( a \\).\n\nHere, \\( p = 97 \\), so \\( 79^{96} \\equiv 1 \\pmod{97} \\). But I need \\( 79^8 \\), which is not directly helpful. Maybe I can compute \\( 79^2 \\), then \\( 79^4 \\), and so on.\n\nFirst, compute \\( 79^2 \\pmod{97} \\):\n\n\\( 79^2 = 6241 \\). Now, divide 6241 by 97.\n\n97 times 64 is 6208, so 6241 - 6208 = 33. Therefore, \\( 79^2 \\equiv 33 \\pmod{97} \\).\n\nNext, \\( 79^4 = (79^2)^2 \\equiv 33^2 \\pmod{97} \\).\n\n\\( 33^2 = 1089 \\). 97 times 11 is 1067, so 1089 - 1067 = 22. Therefore, \\( 79^4 \\equiv 22 \\pmod{97} \\).\n\nThen, \\( 79^8 = (79^4)^2 \\equiv 22^2 \\pmod{97} \\).\n\n\\( 22^2 = 484 \\). 97 times 4 is 388, so 484 - 388 = 96. Therefore, \\( 79^8 \\equiv 96 \\pmod{97} \\).\n\nThus, \\( 2019^8 + 1 \\equiv 96 + 1 \\equiv 97 \\equiv 0 \\pmod{97} \\).\n\nSo, 97 divides \\( 2019^8 + 1 \\). Since 97 is the smallest prime congruent to 1 modulo 16 that we've checked and it divides \\( 2019^8 + 1 \\), it must be the least odd prime factor of \\( 2019^8 + 1 \\).\n\nTherefore, the least odd prime factor is 97.\n\n**Final Answer**\n\n\\[ \\boxed{97} \\]\n",
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Raw Rating Output

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}