Find the last three digits of $9^{105}.$
We can write $9^{105} = (10 - 1)^{105}.$  Then by the Binomial Theorem,
\[(10 - 1)^{105} = 10^{105} - \binom{105}{1} 10^{104} + \binom{105}{2} 10^{103} - \dots + \binom{105}{102} 10^3 - \binom{105}{103} 10^2 + \binom{105}{104} 10 - 1.\]All the terms up to $\binom{105}{102} 10^3$ are divisible by $10^3,$ so for the purpose of finding the last three digits, we can ignore them.  We are left with
\begin{align*}
-\binom{105}{103} 10^2 + \binom{105}{104} 10 - 1 &= -\binom{105}{2} 10^2 + \binom{105}{1} 10 - 1 \\
&= -\frac{105 \cdot 104}{2} \cdot 10^2 + 105 \cdot 10 - 1 \\
&= -546000 + 1050 - 1 \\
&= -546000 + 1049.
\end{align*}Hence, the last three digits are $\boxed{049}.$
049
Level 3
Intermediate Algebra
Problem: Find the last three digits of $9^{105}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the last three digits of \(9^{105}\), we need to compute \(9^{105} \mod 1000\). 

First, we can express \(9\) as \(3^2\), so \(9^{105} = (3^2)^{105} = 3^{210}\). 

Next, we will use the Chinese Remainder Theorem (CRT) to simplify our calculations by finding \(3^{210} \mod 8\) and \(3^{210} \mod 125\), since \(1000 = 8 \times 125\).

Calculating \(3^{210} \mod 8\):
Since \(3 \equiv 3 \mod 8\), we can compute the powers of \(3\) modulo \(8\):
- \(3^1 \equiv 3\)
- \(3^2 \equiv 1\)

The powers of \(3\) modulo \(8\) cycle every \(2\) terms. Thus, \(3^{210} = (3^2)^{105} \equiv 1^{105} \equiv 1 \mod 8\).

Now, calculating \(3^{210} \mod 125\):
Using Euler's theorem, we first find \(\phi(125)\):
\(\phi(125) = 125 \cdot (1 - \frac{1}{5}) = 100\).

Since \(3\) and \(125\) are coprime, we can apply Euler's theorem:
\(3^{100} \equiv 1 \mod 125\).

Now, we reduce \(210\) modulo \(100\):
\(210 \mod 100 = 10\).

Thus, \(3^{210} \equiv 3^{10} \mod 125\). We calculate \(3^{10}\):
- \(3^2 = 9\)
- \(3^4 = 9^2 = 81\)
- \(3^8 = 81^2 = 6561\)

Now, we compute \(6561 \mod 125\):
\(6561 \div 125 = 52.488\), so \(52 \times 125 = 6500\).
Thus, \(6561 - 6500 = 61\), so \(3^8 \equiv 61 \mod 125\).

Next, we compute \(3^{10} = 3^8 \cdot 3^2 \equiv 61 \cdot 9 \mod 125\):
Calculating \(61 \cdot 9 = 549\).
Now, we find \(549 \mod 125\):
\(549 \div 125 = 4.392\), so \(4 \times 125 = 500\).
Thus, \(549 - 500 = 49\), so \(3^{10} \equiv 49 \mod 125\).

Now we have:
1. \(3^{210} \equiv 1 \mod 8\)
2. \(3^{210} \equiv 49 \mod 125\)

We can now use the CRT to combine these results. We need to solve the system:
- \(x \equiv 1 \mod 8\)
- \(x \equiv 49 \mod 125\)

Let \(x = 125k + 49\) for some integer \(k\). Substituting into the first congruence:
\(125k + 49 \equiv 1 \mod 8\).

Calculating \(125 \mod 8\):
\(125 \div 8 = 15.625\), so \(15 \times 8 = 120\).
Thus, \(125 \equiv 5 \mod 8\).

Now substituting:
\(5k + 49 \equiv 1 \mod 8\).
Calculating \(49 \mod 8\):
\(49 \div 8 = 6.125\), so \(6 \times 8 = 48\).
Thus, \(49 \equiv 1 \mod 8\).

Now we have:
\(5k + 1 \equiv 1 \mod 8\) simplifies to \(5k \equiv 0 \mod 8\).

The multiplicative inverse of \(5\) modulo \(8\) is \(5\) (since \(5 \cdot 5 = 25 \equiv 1 \mod 8\)). Thus, multiplying both sides by \(5\):
\(k \equiv 0 \mod 8\).

Let \(k = 8m\) for some integer \(m\). Substituting back:
\(x = 125(8m) + 49 = 1000m + 49\).

Thus, \(x \equiv 49 \mod 1000\).

Therefore, the last three digits of \(9^{105}\) are \(049\).<|END_OF_LATENT|> The final answer is $049$.