Okay, so I need to find the remainder when 129 raised to the 34th power plus 96 raised to the 38th power is divided by 11. Hmm, that sounds like a modular arithmetic problem. Let me recall how modular arithmetic works here. When dealing with exponents modulo a number, it's often helpful to first reduce the base modulo that number. That should simplify things a bit.

Let me start by finding 129 mod 11 and 96 mod 11. That way, I can work with smaller numbers instead of dealing with 129 and 96 directly. Let's do that first.

Starting with 129 divided by 11. Well, 11 times 11 is 121, right? So 129 minus 121 is 8. So 129 mod 11 is 8. Similarly, for 96 divided by 11. 11 times 8 is 88, so 96 minus 88 is 8. Therefore, 96 mod 11 is also 8. Wait, both bases reduce to 8 modulo 11? That's interesting. So then, the problem simplifies to finding 8^34 + 8^38 mod 11. That might be easier.

Now, I need to compute 8^34 mod 11 and 8^38 mod 11, then add them together and take mod 11 again. But calculating 8 raised to such high exponents directly seems impractical. Maybe there's a pattern or a theorem I can use here. Oh, right! Fermat's Little Theorem. It states that if p is a prime number and a is not divisible by p, then a^(p-1) ≡ 1 mod p. Since 11 is prime and 8 is not divisible by 11, Fermat's Little Theorem applies here. So 8^10 ≡ 1 mod 11. That could help break down the exponents into smaller parts.

Let me verify that. 10 is 11-1, so yes, according to Fermat's Little Theorem, 8^10 ≡ 1 mod 11. So any exponent can be reduced modulo 10 when dealing with powers of 8 modulo 11. For example, 8^k ≡ 8^(k mod 10) mod 11. Let me check that with a smaller exponent. Let's take 8^2. 8^2 is 64. 64 divided by 11 is 5*11=55, remainder 9. 8^2 mod 11 is 9. Now, 8^(2 mod 10) is 8^2, which is 64 mod 11 is 9. That works. Let's try 8^3. 8^3 is 512. 512 divided by 11 is 46*11=506, remainder 6. 8^3 mod 11 is 6. 8^(3 mod 10) is 8^3, which is 512 mod 11 is 6. That also works. So the theorem holds here. Therefore, I can use this to reduce the exponents.

So for 8^34 mod 11, since 34 divided by 10 is 3 with a remainder of 4. So 34 mod 10 is 4. Therefore, 8^34 ≡ 8^4 mod 11. Similarly, 8^38 mod 11: 38 divided by 10 is 3 with a remainder of 8. So 38 mod 10 is 8. Therefore, 8^38 ≡ 8^8 mod 11.

Now, I need to compute 8^4 mod 11 and 8^8 mod 11. Let's compute them step by step.

First, 8^1 mod 11 is 8. 8^2 mod 11 is 64 mod 11. As before, 64 divided by 11 is 5*11=55, remainder 9. So 8^2 ≡ 9 mod 11.

Then, 8^3 = 8^2 * 8. So 9 * 8 = 72. 72 divided by 11 is 6*11=66, remainder 6. So 8^3 ≡ 6 mod 11.

8^4 = 8^3 * 8. 6*8 = 48. 48 divided by 11 is 4*11=44, remainder 4. So 8^4 ≡ 4 mod 11.

So, 8^34 ≡ 8^4 ≡ 4 mod 11.

Now, let's compute 8^8 mod 11. Maybe I can build up from 8^4. Since 8^4 is 4 mod 11, then 8^8 is (8^4)^2 mod 11. So 4 squared is 16. 16 mod 11 is 5. Therefore, 8^8 ≡ 5 mod 11.

So, 8^38 ≡ 8^8 ≡ 5 mod 11.

Therefore, combining both results: 8^34 + 8^38 ≡ 4 + 5 ≡ 9 mod 11.

Therefore, the remainder when 129^34 + 96^38 is divided by 11 is 9.

Wait, let me verify this again to make sure I didn't make any mistakes. Let's go through the steps again.

First, 129 mod 11: 11*11=121, 129-121=8. Correct. 96 mod 11: 11*8=88, 96-88=8. Correct. So both bases are 8 mod 11. Then, using Fermat's Little Theorem, 8^10 ≡ 1 mod 11. Therefore, exponents can be reduced modulo 10.

34 divided by 10 is 3 with remainder 4. So 8^34 ≡ 8^4 mod 11. 8^4 is 8*8*8*8. Let's compute 8^2 first: 64 mod 11=9. Then 8^4=(8^2)^2=9^2=81 mod 11. 81 divided by 11 is 7*11=77, remainder 4. So 8^4 ≡4 mod 11. Correct.

Similarly, 38 divided by 10 is 3 with remainder 8. So 8^38 ≡8^8 mod 11. 8^8 is (8^4)^2. Since 8^4 ≡4 mod 11, then 4^2=16 mod 11=5. So 8^8 ≡5 mod 11. Correct.

Then, adding 4 +5=9 mod 11. So the remainder is 9. That seems correct.

Alternatively, maybe I can check with another approach. Let's compute the exponents step by step without using Fermat's Little Theorem to confirm.

Starting with 8^1=8 mod11=8

8^2=8*8=64 mod11=9

8^3=8^2*8=9*8=72 mod11=72-66=6

8^4=8^3*8=6*8=48 mod11=48-44=4

8^5=8^4*8=4*8=32 mod11=32-22=10

8^6=8^5*8=10*8=80 mod11=80-77=3

8^7=8^6*8=3*8=24 mod11=24-22=2

8^8=8^7*8=2*8=16 mod11=5

Yes, so 8^8 ≡5 mod11. Then 8^9=8^8*8=5*8=40 mod11=40-33=7

8^10=8^9*8=7*8=56 mod11=56-55=1. So 8^10≡1 mod11. Which matches Fermat's Little Theorem.

So, since 8^10≡1, then 8^34=8^(10*3 +4)= (8^10)^3 *8^4 ≡1^3 *4=4 mod11. Similarly, 8^38=8^(10*3 +8)= (8^10)^3 *8^8 ≡1^3 *5=5 mod11. Then 4+5=9 mod11. So that's consistent.

Therefore, the remainder is 9. So I think that's the correct answer. Let me see if there's another way to check.

Alternatively, maybe calculate 8^34 and 8^38 mod11 separately using patterns.

Looking at the powers of 8 mod11:

8^1=8

8^2=9

8^3=6

8^4=4

8^5=10

8^6=3

8^7=2

8^8=5

8^9=7

8^10=1

Then the cycle repeats every 10 exponents. So the cycle length is 10. Therefore, exponents cycle every 10. So 8^k mod11 cycles every 10. Therefore, 8^34: 34 divided by 10 is 3 remainder 4, so 8^4=4. 8^38: 38 divided by 10 is 3 remainder 8, so 8^8=5. Then 4+5=9. Same result.

Alternatively, maybe using another theorem. Wait, maybe using the Chinese Remainder Theorem? But since 11 is prime, maybe not necessary. Alternatively, since 8 and 11 are coprime, Euler's theorem applies, which is essentially Fermat's Little Theorem for modulus 11. So that's the same as before.

Alternatively, maybe compute the exponents directly using smaller cycles. But since 8^10≡1, the cycle is 10, so exponents can be reduced modulo 10. So that's the key here. So the plan is solid.

Therefore, I think the remainder is 9. So the final answer is 9. Let me just confirm once more.

Compute 129^34 mod11: 129≡8 mod11, so 8^34≡4 mod11.

96^38 mod11: 96≡8 mod11, so 8^38≡5 mod11.

4 +5=9. 9 mod11=9. Yes, correct. So remainder is 9.

I think that's thorough. I considered multiple approaches, checked the cycles, applied Fermat's Little Theorem, and verified the calculations step by step. All methods converge to the same answer. Therefore, I'm confident the remainder is 9.
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