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Okay, so I need to figure out how to solve this problem where I have to find the value of 3 raised to the power of 2021 modulo 7. Hmm, modular arithmetic, right? I remember that when dealing with exponents modulo a number, there might be a pattern or cycle in the remainders. Let me think.




First, maybe I should recall some basic properties of exponents and modulus. I know that if I can find a cycle length for the powers of 3 modulo 7, then I can reduce the exponent 2021 modulo that cycle length to simplify the problem. That sounds like a plan.


Let me start by calculating the first few powers of 3 modulo 7 to see if there's a repeating pattern.


3^1 = 3 mod 7 is 3.


3^2 = 9 mod 7. Since 9 divided by 7 is 1 with a remainder of 2, so 9 mod 7 is 2.


3^3 = 3^2 * 3 = 2 * 3 = 6 mod 7. That's 6.


3^4 = 3^3 * 3 = 6 * 3 = 18 mod 7. 18 divided by 7 is 2 with a remainder of 4, so 18 mod 7 is 4.


3^5 = 3^4 * 3 = 4 * 3 = 12 mod 7. 12 divided by 7 is 1 with a remainder of 5, so 12 mod 7 is 5.


3^6 = 3^5 * 3 = 5 * 3 = 15 mod 7. 15 divided by 7 is 2 with a remainder of 1, so 15 mod 7 is 1.


Wait a minute, 3^6 mod 7 is 1? That's interesting. So 3^6 ≡ 1 mod 7. That means the cycle length here is 6. Because after 6 exponents, the remainder cycles back to 1. So the powers of 3 modulo 7 repeat every 6 steps. So the pattern is 3, 2, 6, 4, 5, 1, and then repeats.


So if the cycle length is 6, then 3^(6k) ≡ 1 mod 7 for any integer k. Therefore, 3^(6k + r) ≡ 3^r * (3^6)^k ≡ 3^r * 1^k ≡ 3^r mod 7. So to compute 3^2021 mod 7, I need to find the remainder when 2021 is divided by 6. That will give me the exponent r, and then I can compute 3^r mod 7.


Alright, let's compute 2021 divided by 6. Let me do that division. 6 goes into 2021 how many times?


First, 6*300 = 1800. 2021 - 1800 = 221.


6*30 = 180. 221 - 180 = 41.


6*6 = 36. 41 - 36 = 5.


So total is 300 + 30 + 6 = 336, with a remainder of 5. So 2021 = 6*336 + 5. Therefore, 2021 mod 6 is 5.


Therefore, 3^2021 mod 7 is the same as 3^5 mod 7. From my earlier calculations, 3^5 mod 7 is 5. So the answer should be 5.


Wait, let me verify that again. Let me check the steps once more to make sure I didn't make a mistake.


First, calculating the cycle:


3^1 mod7=3


3^2=9 mod7=2


3^3=6 mod7=6


3^4=18 mod7=4


3^5=12 mod7=5


3^6=15 mod7=1


Yes, cycle length is 6. Then 2021 divided by 6. Let me check that division again.


2021 divided by 6:


6336=2016. 2021-2016=5. So remainder 5. So exponent 2021 is 6336 +5, so 3^2021= (3^6)^336 *3^5. Since 3^6 ≡1 mod7, this is 1^336 *3^5 ≡3^5 mod7. And 3^5 mod7=5. So yes, the answer is 5.


Alternatively, maybe I can use Fermat's little theorem. Fermat's little theorem states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) ≡1 mod p. Here, 7 is prime, and 3 is not divisible by 7, so 3^6 ≡1 mod7. Which is exactly what we found earlier. So that's another way to confirm the cycle length is 6.


Therefore, using Fermat's little theorem, 3^6 ≡1 mod7, so 3^(6k)≡1^k=1 mod7. Then 3^2021=3^(6*336 +5)= (3^6)^336 *3^5 ≡1^336 3^5 ≡3^5 mod7. And 3^5=243. 243 divided by 7: 734=238, remainder 5. So 243 mod7=5. Therefore, 3^2021 mod7=5.


So both methods lead to the same result. Therefore, I think the answer is 5.


But just to be absolutely thorough, let me compute 3^5 mod7 again. 3^1=3, 3^2=9-7=2, 3^3=23=6, 3^4=63=18-14=4, 3^5=4*3=12-7=5. Yep, that's 5. So that's correct.


Alternatively, maybe I can write out the exponents in a table to see the pattern:


Exponent | Result | Mod 7


1 | 3 | 3


2 | 9 | 2


3 | 27 | 6


4 | 81 | 4


5 | 243 | 5


6 | 729 | 1


7 | 2187 | 3 (since 7293=2187; 2187 mod7: 7312=2184, remainder 3)


8 | 6561 | 2 (3*3=9 mod7=2)


So yes, the cycle repeats every 6 exponents. So exponent 7 is same as 1, exponent 8 same as 2, etc. Therefore, exponent 2021: since 2021 divided by 6 gives remainder 5, exponent 5 corresponds to result 5 mod7. So that's consistent.


Therefore, I can be confident that the answer is 5.


Final Answer
The value of (3^{2021} \mod 7) is \boxed{5}.




To find the value of (3^{2021} \mod 7), we observe the repeating pattern of the powers of 3 modulo 7.


First, we calculate the first few powers of 3 modulo 7:
[
\begin{align*}
3^1 &\equiv 3 \mod 7, \
3^2 &\equiv 9 \equiv 2 \mod 7, \
3^3 &\equiv 3 \cdot 2 = 6 \mod 7, \
3^4 &\equiv 3 \cdot 6 = 18 \equiv 4 \mod 7, \
3^5 &\equiv 3 \cdot 4 = 12 \equiv 5 \mod 7, \
3^6 &\equiv 3 \cdot 5 = 15 \equiv 1 \mod 7.
\end{align*}
]
We see that (3^6 \equiv 1 \mod 7), indicating a cycle length of 6.


Next, we reduce the exponent 2021 modulo 6 to find the position within the cycle:
[
2021 \div 6 = 336 \text{ remainder } 5.
]
Thus, (2021 \equiv 5 \mod 6). Therefore, (3^{2021} \equiv 3^5 \mod 7).


From our earlier calculations, we know (3^5 \equiv 5 \mod 7).


Hence, the value of (3^{2021} \mod 7) is (\boxed{5}).