The question asks to prove that $11^{10} - 1$ is divisible by 100.

Here's a solution:

Solution:

We need to show that $11^{10} - 1 \equiv 0 \pmod{100}$.

To do this, we can use modular arithmetic and the Chinese Remainder Theorem, by showing that:

1. $11^{10} - 1 \equiv 0 \pmod{4}$
2. $11^{10} - 1 \equiv 0 \pmod{25}$

Step 1: Check modulo 4

Notice that $11 \equiv 3 \pmod{4}$. Therefore: $11^{10} \equiv 3^{10} \pmod{4}.$ Calculating powers of 3 modulo 4, we observe that: $3^2 \equiv 1 \pmod{4}.$ Thus, $3^{10} = (3^2)^5 \equiv 1^5 = 1 \pmod{4}$. This gives: $11^{10} \equiv 1 \pmod{4}.$ Then $11^{10} - 1 \equiv 1 - 1 = 0 \pmod{4}$.

Step 2: Check modulo 25

Similarly, we note that $11 \equiv 11 \pmod{25}$. Using Euler's theorem, since 11 and 25 are coprime, we know that: $11^{20} \equiv 1 \pmod{25}.$ Thus, we can calculate $11^{10}$ modulo 25 by direct computation or by further breaking down the powers: $11^2 = 121 \equiv 21 \pmod{25},$ $11^4 = 21^2 = 441 \equiv 16 \pmod{25},$ $11^8 = 16^2 = 256 \equiv 6 \pmod{25},$ $11^{10} = 11^8 \cdot 11^2 \equiv 6 \cdot 21 = 126 \equiv 1 \pmod{25}.$ Therefore: $11^{10} - 1 \equiv 1 - 1 = 0 \pmod{25}.$

Conclusion:

Since $11^{10} - 1 \equiv 0 \pmod{4}$ and $11^{10} - 1 \equiv 0 \pmod{25}$, by the Chinese Remainder Theorem, we conclude that: $11^{10} - 1 \equiv 0 \pmod{100}.$ This proves that $11^{10} - 1$ is divisible by 100.

Would you like more details on any specific part of this proof?

Here are five follow-up questions you might find interesting:

1. How does the Chinese Remainder Theorem work, and why is it applicable here?
2. Why is Euler's theorem useful for computing powers in modular arithmetic?
3. Can you apply a similar method to prove that $13^{10} - 1$ is divisible by 100?
4. How can modular arithmetic be used to simplify computations in larger powers?
5. What other techniques could be used to prove divisibility in similar problems?

Tip: Remember that simplifying powers using modular properties can make large calculations much more manageable!