To find the tens digit of $7^{2011}$, we can use modular arithmetic to reduce the problem.

Step-by-Step Solution

1. Find the periodicity of powers of 7 modulo 100.

   The goal is to find $7^{2011} \mod 100$, and the tens digit will be the second last digit of this result.

   Since powers of 7 repeat in cycles modulo 100, let's first identify the cycle for powers of 7 modulo 100.

   We begin by calculating the first few powers of 7 modulo 100:

   $7^1 \mod 100 = 7$ $7^2 \mod 100 = 49$ $7^3 \mod 100 = 343 \mod 100 = 43$ $7^4 \mod 100 = 7 \times 43 = 301 \mod 100 = 1$

   This shows that the powers of 7 modulo 100 repeat every 4 terms, specifically the cycle is $7, 49, 43, 1$.

2. Find where $7^{2011}$ falls in the cycle.

   Since the powers of 7 modulo 100 repeat every 4 terms, we need to determine the position of $2011$ in this cycle. This can be done by calculating:

   $2011 \mod 4$

   Dividing 2011 by 4 gives a remainder of 3, so $7^{2011} \mod 100$ corresponds to $7^3 \mod 100$.

3. Conclusion:

   From the earlier calculations, $7^3 \mod 100 = 43$. Therefore, the last two digits of $7^{2011}$ are 43, and the tens digit is:

   $\boxed{4}$

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Would you like more details on any part of the process, or do you have further questions?

Related Questions:

1. What is the units digit of $7^{2011}$?
2. How do you find the last digit of powers of numbers like 3, 4, or 5?
3. What is the periodicity of powers of numbers modulo 100 in general?
4. How does Euler’s Theorem help in finding large powers modulo a number?
5. Can you calculate the tens digit of $11^{2011}$?

Tip: Always start by identifying the pattern or cycle of powers when working with modular arithmetic, as it can save time and effort in large exponentiation problems.