Question

Find the smallest 5-digit number exactly divisible by 24 and 36.

• A. 10008
• B. 10080
• C. 10024
• D. 10036

Hint:

To quickly multiply numbers like \(139 \times 72\) in your head or on scratch paper, rewrite \(139\) as \((140 - 1)\):
\[ (140 - 1) \times 72 = 140 \times 72 - 72 \]
\[ = 10080 - 72 = 10008 \]
This technique is highly reliable and prevents basic arithmetic errors under exam pressure.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires us to find the smallest 5-digit integer that can be divided by both 24 and 36 without leaving any remainder.

Step 2: Key Formula or Approach:
1. Any number exactly divisible by both 24 and 36 must be a multiple of their Lowest Common Multiple (L.C.M.).
2. Therefore, we must first calculate the L.C.M. of 24 and 36.
3. Once the L.C.M. is found, we identify the smallest 5-digit number (which is 10,000) and find the smallest multiple of the L.C.M. that is greater than or equal to 10,000.

Step 3: Detailed Explanation:
1. Find the prime factorization of both numbers:
\[ 24 = 2^3 \times 3^1 \]
\[ 36 = 2^2 \times 3^2 \]
2. Determine the L.C.M. by taking the highest power of all prime factors involved:
\[ \text{L.C.M.}(24, 36) = 2^3 \times 3^2 = 8 \times 9 = 72 \]
3. The required number must be a multiple of 72.
4. The smallest 5-digit number is 10,000. Let us divide 10,000 by 72 to see where the multiples lie:
\[ \frac{10000}{72} \approx 138.89 \]
5. The smallest integer multiple of 72 that is a 5-digit number is given by multiplying 72 by the next higher integer, which is 139:
\[ \text{Required Number} = 139 \times 72 \]
6. Perform the multiplication:
\[ 139 \times 72 = 10008 \]
7. Let us double-check if 10,008 is divisible by both 24 and 36:
\[ \frac{10008}{24} = 417 \]
\[ \frac{10008}{36} = 278 \]
Both divisions result in integers with no remainder.

Step 4: Final Answer:
The smallest 5-digit number exactly divisible by 24 and 36 is 10008, which corresponds to option (A).