Question

The smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively is

• A. 224
• B. 244
• C. 204
• D. 214

Answer 1

The Correct Option is C

Step 1: Express the problem mathematically.

Let the number be \( x \). Then:

\[ x \equiv 8 \pmod{28}, \quad x \equiv 12 \pmod{32}. \]

Step 2: Rewrite the congruences.

We can rewrite the congruences as:

\[ x = 28k + 8 \quad \text{(for some integer } k\text{)}, \quad x = 32m + 12 \quad \text{(for some integer } m\text{)}. \]

Step 3: Find the least common multiple (LCM) of 28 and 32.

The LCM of 28 and 32 is:

\[ \text{LCM}(28, 32) = 2^5 \times 7 = 224. \]

Step 4: Solve for the smallest \( x \).

The general solution is:

\[ x = \text{LCM}(28, 32) + r, \]

where \( r \) satisfies both congruences. Testing values, we find:

\[ x = 224 - 20 = 204. \]

Final Answer: The smallest number is \( \mathbf{204} \), which corresponds to option \( \mathbf{(3)} \).