To find the last two digits of $9^{2019}$, we need to calculate $9^{2019} \mod 100$. We can solve this using the Chinese Remainder Theorem by separately calculating:

1. $9^{2019} \mod 4$
2. $9^{2019} \mod 25$

Step 1: $9^{2019} \mod 4$

Since $9 \equiv 1 \mod 4$, we have:

$9^{2019} \equiv 1^{2019} \equiv 1 \mod 4$

Step 2: $9^{2019} \mod 25$

To calculate $9^{2019} \mod 25$, we can use Euler's theorem. The Euler's totient function $\phi(25) = 20$, so by Euler's theorem:

$9^{20} \equiv 1 \mod 25$

Now, since $2019 \mod 20 = 19$, we need to calculate $9^{19} \mod 25$.

We can compute powers of 9 modulo 25:

• $9^2 = 81 \equiv 6 \mod 25$
• $9^4 = (9^2)^2 = 6^2 = 36 \equiv 11 \mod 25$
• $9^8 = (9^4)^2 = 11^2 = 121 \equiv 21 \mod 25$
• $9^{16} = (9^8)^2 = 21^2 = 441 \equiv 16 \mod 25$

Now, break $9^{19} = 9^{16} \times 9^3 \mod 25$:

• $9^3 = 9 \times 9^2 = 9 \times 6 = 54 \equiv 4 \mod 25$
• $9^{19} = 9^{16} \times 9^3 = 16 \times 4 = 64 \equiv 14 \mod 25$

Step 3: Combine using the Chinese Remainder Theorem

We now have the system of congruences:

$x \equiv 1 \mod 4$ $x \equiv 14 \mod 25$

Using the method of successive substitution or the Chinese Remainder Theorem, solve for $x$. Let $x = 25k + 14$, and substitute into the first congruence:

$25k + 14 \equiv 1 \mod 4$

Since $25 \equiv 1 \mod 4$ and $14 \equiv 2 \mod 4$, the equation becomes:

$k + 2 \equiv 1 \mod 4$

This gives $k \equiv 3 \mod 4$, so $k = 4m + 3$. Substituting this into $x = 25k + 14$, we get:

$x = 25(4m + 3) + 14 = 100m + 89$

Thus, $x \equiv 89 \mod 100$.

The last two digits of $9^{2019}$ are $\boxed{89}$.

Would you like more details, or do you have any questions?

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Here are 5 related questions you might find interesting:

1. How can the Chinese Remainder Theorem be applied in different modular arithmetic problems?
2. What is Euler's theorem, and how is it useful in solving problems involving large powers modulo numbers?
3. How do you calculate large powers mod prime numbers?
4. What are efficient methods to calculate $9^{2019} \mod 25$ using modular exponentiation?
5. How does the method of successive substitution work in solving modular systems?

Tip: When dealing with powers in modular arithmetic, break down the problem into smaller steps and apply the Chinese Remainder Theorem if the modulus can be factored into coprimes!