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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} ]

So I've got this equation: 1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9 = 479, and it's supposed to be incorrect, but I need to add just one pair of parentheses to make it true. Hmm, let's see.

First, I should probably understand what the equation is currently doing. In math, multiplication is performed before addition according to the order of operations (PEMDAS/BODMAS rules). So, in the expression 1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9, the multiplications would be calculated first, and then the additions.

Let's compute it step by step without any parentheses:

First, the multiplications:

2 * 3 = 6

4 * 5 = 20

6 * 7 = 42

8 * 9 = 72

Now, plug these back into the equation:

1 + 6 + 20 + 42 + 72

Now, add them up:

1 + 6 = 7

7 + 20 = 27

27 + 42 = 69

69 + 72 = 141

So, without any parentheses, the equation equals 141, not 479. So, clearly, something needs to change.

Now, I need to add one pair of parentheses to make this equal to 479. Parentheses can change the order of operations, so whatever is inside the parentheses gets calculated first.

I need to think about where to place these parentheses to alter the result significantly, since 141 is quite a bit smaller than 479.

Maybe I can group some additions and multiplications in a way that increases the result.

Let me consider different placements:

Option 1: (1 + 2) * 3 + 4 * 5 + 6 * 7 + 8 * 9

Calculate inside the parentheses first:

1 + 2 = 3

Then, 3 * 3 = 9

Now, the equation is: 9 + 4 * 5 + 6 * 7 + 8 * 9

Next, multiplications:

4 * 5 = 20

6 * 7 = 42

8 * 9 = 72

So, 9 + 20 + 42 + 72

Add them up:

9 + 20 = 29

29 + 42 = 71

71 + 72 = 143

Still not 479.

Option 2: 1 + (2 * 3 + 4) * 5 + 6 * 7 + 8 * 9

First, inside the parentheses:

2 * 3 = 6

6 + 4 = 10

Now, multiply by 5: 10 * 5 = 50

So, the equation becomes: 1 + 50 + 6 * 7 + 8 * 9

Next, multiplications:

6 * 7 = 42

8 * 9 = 72

Now, add them up:

1 + 50 = 51

51 + 42 = 93

93 + 72 = 165

Still not 479.

Option 3: 1 + 2 * (3 + 4) * 5 + 6 * 7 + 8 * 9

First, inside the parentheses:

3 + 4 = 7

Then, 2 * 7 = 14

Then, 14 * 5 = 70

So, the equation is: 1 + 70 + 6 * 7 + 8 * 9

Next, multiplications:

6 * 7 = 42

8 * 9 = 72

Now, add them up:

1 + 70 = 71

71 + 42 = 113

113 + 72 = 185

Still not 479.

Option 4: 1 + 2 * 3 + 4 * (5 + 6) * 7 + 8 * 9

First, inside the parentheses:

5 + 6 = 11

Then, 11 * 7 = 77

So, the equation is: 1 + 2 * 3 + 4 * 77 + 8 * 9

Next, multiplications:

2 * 3 = 6

4 * 77 = 308

8 * 9 = 72

Now, add them up:

1 + 6 = 7

7 + 308 = 315

315 + 72 = 387

Closer, but still not 479.

Option 5: 1 + 2 * 3 + 4 * 5 + 6 * (7 + 8) * 9

First, inside the parentheses:

7 + 8 = 15

Then, 15 * 9 = 135

So, the equation is: 1 + 2 * 3 + 4 * 5 + 6 * 135

Next, multiplications:

2 * 3 = 6

4 * 5 = 20

6 * 135 = 810

Now, add them up:

1 + 6 = 7

7 + 20 = 27

27 + 810 = 837

Way too big.

Hmm, 837 is much larger than 479. Maybe I need to reduce the impact of the parentheses.

Option 6: 1 + 2 * (3 + 4 * 5) + 6 * 7 + 8 * 9

First, inside the parentheses:

3 + 4 * 5

Wait, in this case, multiplication comes before addition inside the parentheses, so 4 * 5 = 20, then 3 + 20 = 23

Then, 2 * 23 = 46

So, the equation is: 1 + 46 + 6 * 7 + 8 * 9

Next, multiplications:

6 * 7 = 42

8 * 9 = 72

Now, add them up:

1 + 46 = 47

47 + 42 = 89

89 + 72 = 161

Still not 479.

Option 7: 1 + 2 * 3 + 4 * (5 + 6 * 7) + 8 * 9

First, inside the parentheses:

5 + 6 * 7

Multiplication first: 6 * 7 = 42, then 5 + 42 = 47

Then, 4 * 47 = 188

So, the equation is: 1 + 2 * 3 + 188 + 8 * 9

Next, multiplications:

2 * 3 = 6

8 * 9 = 72

Now, add them up:

1 + 6 = 7

7 + 188 = 195

195 + 72 = 267

Still not 479.

Option 8: 1 + 2 * 3 + 4 * 5 + 6 * (7 + 8 * 9)

First, inside the parentheses:

7 + 8 * 9

Multiplication first: 8 * 9 = 72, then 7 + 72 = 79

Then, 6 * 79 = 474

So, the equation is: 1 + 2 * 3 + 4 * 5 + 474

Next, multiplications:

2 * 3 = 6

4 * 5 = 20

Now, add them up:

1 + 6 = 7

7 + 20 = 27

27 + 474 = 479

Perfect! That equals 479.

So, by placing the parentheses around 7 + 8 * 9, and multiplying that sum by 6, we get the desired result.

Therefore, the corrected equation is: 1 + 2 * 3 + 4 * 5 + 6 * (7 + 8 * 9) = 479

Final Answer

[ \boxed{1 + 2 \times 3 + 4 \times 5 + 6 \times (7 + 8 \times 9) = 479} ]