To find all 3-digit numbers which are the sums of the cubes of their digits, we can follow these steps:
### Step 1: Define the 3-digit number
Let the 3-digit number be represented as \( xyz \), where \( x \), \( y \), and \( z \) are the digits of the number. The value of the number can be expressed as:
\[
N = 100x + 10y + z
\]
where \( x \) is the hundreds digit, \( y \) is the tens digit, and \( z \) is the units digit.
### Step 2: Set up the equation
According to the problem, we need to find numbers such that:
\[
N = x^3 + y^3 + z^3
\]
This means:
\[
100x + 10y + z = x^3 + y^3 + z^3
\]
### Step 3: Determine the range of digits
Since \( x \) is the hundreds digit, it must be between 1 and 9 (inclusive). The digits \( y \) and \( z \) can be between 0 and 9 (inclusive). Therefore, the possible values for \( x \), \( y \), and \( z \) are:
- \( x \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
- \( y \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
- \( z \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
### Step 4: Check all combinations
We will check all combinations of \( x \), \( y \), and \( z \) within their respective ranges to see if they satisfy the equation \( 100x + 10y + z = x^3 + y^3 + z^3 \).
### Step 5: Calculate and verify
We can iterate through all possible values of \( x \), \( y \), and \( z \) and calculate \( x^3 + y^3 + z^3 \) for each combination to see if it equals \( 100x + 10y + z \).
1. For \( x = 1 \):
- \( y = 5, z = 3 \): \( 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153 \) (valid)
2. For \( x = 3 \):
- \( y = 7, z = 0 \): \( 3^3 + 7^3 + 0^3 = 27 + 343 + 0 = 370 \) (valid)
3. For \( x = 3 \):
- \( y = 7, z = 1 \): \( 3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371 \) (valid)
4. For \( x = 4 \):
- \( y = 0, z = 7 \): \( 4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407 \) (valid)
### Step 6: List all valid numbers
The valid 3-digit numbers that are equal to the sum of the cubes of their digits are:
- 153
- 370
- 371
- 407
### Final Answer
The 3-digit numbers which are the sums of the cubes of their digits are: **153, 370, 371, and 407.**
---
