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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!

Buktikan bahwa 11^10 - 1 habis dibagi 100.

Discover how to prove that 11^10 - 1 is divisible by 100 using modular arithmetic techniques and the Chinese Remainder Theorem. This solution includes applications of Euler's Theorem and is suitable for high school students studying advanced number theory.

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