If n ends in 3, then `n^3` ends in 7.

Answer

To determine if the statement "If n ends in 3, then n^3 ends in 7" is true or false, we can follow these steps:

### Step 1: Understand the problem
We need to check if a number \( n \) that ends in 3, when cubed (\( n^3 \)), will always end in 7.

### Step 2: Choose a number that ends in 3
Let's take \( n = 3 \) as our example since it is the simplest number that ends in 3.

### Step 3: Calculate \( n^3 \)
Now, we will calculate \( n^3 \):
\[ n^3 = 3^3 = 27 \]

### Step 4: Check the last digit of \( n^3 \)
The last digit of 27 is 7. Therefore, when \( n = 3 \), \( n^3 \) ends in 7.

### Step 5: Generalize the conclusion
To confirm that this holds true for any number ending in 3, we can represent \( n \) as \( n = 10k + 3 \) for some integer \( k \).

### Step 6: Calculate \( n^3 \) in general form
Now, we will calculate \( n^3 \):
\[ n^3 = (10k + 3)^3 \]
Using the binomial expansion:
\[ n^3 = (10k)^3 + 3 \times (10k)^2 \times 3 + 3 \times (10k) \times 3^2 + 3^3 \]
\[ = 1000k^3 + 3 \times 100k^2 \times 3 + 3 \times 10k \times 9 + 27 \]
\[ = 1000k^3 + 900k^2 + 270k + 27 \]

### Step 7: Identify the last digit
The last digit of \( n^3 \) is determined by the last digit of the constant term, which is 27. The last digit of 27 is 7.

### Conclusion
Thus, for any integer \( n \) that ends in 3, \( n^3 \) will always end in 7. Therefore, the statement is **true**.

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