The remainder obtained when `23^3 + 31^3` is divided by 54 :

Options

A. 0
B. 1
C. 3
D. can't be determined

Answer

To find the remainder when \( 23^3 + 31^3 \) is divided by 54, we can use the formula for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

In our case, let \( a = 23 \) and \( b = 31 \).

### Step 1: Calculate \( a + b \)
\[ a + b = 23 + 31 = 54 \]

### Step 2: Calculate \( a^2 \), \( b^2 \), and \( ab \)
\[ a^2 = 23^2 = 529 \]
\[ b^2 = 31^2 = 961 \]
\[ ab = 23 \times 31 = 713 \]

### Step 3: Substitute into the formula
Now substituting into the formula:
\[ a^2 - ab + b^2 = 529 - 713 + 961 \]

### Step 4: Simplify the expression
Calculating the above:
\[ 529 - 713 = -184 \]
\[ -184 + 961 = 777 \]

### Step 5: Substitute back into the sum of cubes formula
Now we have:
\[ 23^3 + 31^3 = (54)(777) \]

### Step 6: Find the remainder when divided by 54
Since \( 23^3 + 31^3 \) is a multiple of 54, the remainder when \( 23^3 + 31^3 \) is divided by 54 is:
\[ \text{Remainder} = 0 \]

Thus, the final answer is:
\[ \text{The remainder obtained when } 23^3 + 31^3 \text{ is divided by } 54 \text{ is } 0. \]

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