Find the last three digits of the number `27^(27)dot`

Explanation

To find the last three digits of the number \( 27^{27} \), we can follow these steps:

### Step 1: Rewrite the base
We start by rewriting \( 27 \) in terms of its prime factors:
\[ 27 = 3^3 \]
Thus, we can express \( 27^{27} \) as:
\[ 27^{27} = (3^3)^{27} = 3^{81} \]

### Step 2: Find \( 3^{81} \mod 1000 \)
To find the last three digits, we need to compute \( 3^{81} \mod 1000 \). We can use the Chinese Remainder Theorem (CRT) by breaking it down into modulo \( 8 \) and modulo \( 125 \).

### Step 3: Calculate \( 3^{81} \mod 8 \)
Calculating \( 3^{81} \mod 8 \):
\[ 3^2 = 9 \equiv 1 \mod 8 \]
Thus, \( 3^{81} = (3^2)^{40} \cdot 3 \equiv 1^{40} \cdot 3 \equiv 3 \mod 8 \).

### Step 4: Calculate \( 3^{81} \mod 125 \)
Now, we calculate \( 3^{81} \mod 125 \) using Euler's theorem. First, we find \( \phi(125) \):
\[ \phi(125) = 125 \left(1 - \frac{1}{5}\right) = 125 \cdot \frac{4}{5} = 100 \]
Since \( 81 < 100 \), we can directly compute \( 3^{81} \mod 125 \).

We can calculate powers of \( 3 \) modulo \( 125 \):
\[ 3^1 = 3 \]
\[ 3^2 = 9 \]
\[ 3^4 = 81 \]
\[ 3^5 = 243 \equiv 118 \mod 125 \]
\[ 3^{10} = (3^5)^2 = 118^2 = 13924 \equiv 74 \mod 125 \]
\[ 3^{20} = (3^{10})^2 = 74^2 = 5476 \equiv 101 \mod 125 \]
\[ 3^{40} = (3^{20})^2 = 101^2 = 10201 \equiv 76 \mod 125 \]
Now, we can compute \( 3^{80} = (3^{40})^2 \):
\[ 3^{80} = 76^2 = 5776 \equiv 26 \mod 125 \]
Finally, we compute \( 3^{81} = 3^{80} \cdot 3 \):
\[ 3^{81} \equiv 26 \cdot 3 = 78 \mod 125 \]

### Step 5: Solve the system of congruences
Now we have the two congruences:
\[ 3^{81} \equiv 3 \mod 8 \]
\[ 3^{81} \equiv 78 \mod 125 \]

We can use the method of successive substitutions to solve these:
Let \( x \equiv 78 \mod 125 \). We can express \( x \) as:
\[ x = 125k + 78 \]
Substituting into the first congruence:
\[ 125k + 78 \equiv 3 \mod 8 \]
Calculating \( 125 \mod 8 \):
\[ 125 \equiv 5 \mod 8 \]
Thus:
\[ 5k + 78 \equiv 3 \mod 8 \]
Calculating \( 78 \mod 8 \):
\[ 78 \equiv 6 \mod 8 \]
So we have:
\[ 5k + 6 \equiv 3 \mod 8 \implies 5k \equiv -3 \equiv 5 \mod 8 \]
Multiplying both sides by the modular inverse of \( 5 \mod 8 \), which is \( 5 \) (since \( 5 \cdot 5 \equiv 1 \mod 8 \)):
\[ k \equiv 5 \cdot 5 \equiv 25 \equiv 1 \mod 8 \]
Thus, \( k = 8m + 1 \) for some integer \( m \). Substituting back:
\[ x = 125(8m + 1) + 78 = 1000m + 203 \]
Thus:
\[ x \equiv 203 \mod 1000 \]

### Final Answer
The last three digits of \( 27^{27} \) are:
\[ \boxed{203} \]