Question

How many numbers divisible by 132:
264, 396, 462, 792, 968, 2178, 5184, 6336

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

To check divisibility quickly, divide the number mentally by the divisor and see whether the result is a whole number.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find how many numbers in the given list are exactly divisible by \(132\).
Given numbers: \[ 264,\ 396,\ 462,\ 792,\ 968,\ 2178,\ 5184,\ 6336 \]

Step 2: Key Formula or Approach:
A number is divisible by \(132\) if: \[ \frac{\text{Number}}{132} \] gives a whole number (integer).

Step 3: Detailed Explanation:
Check each number one by one: \[ 264 \div 132 = 2 \] Divisible.
\[ 396 \div 132 = 3 \] Divisible.
\[ 462 \div 132 = 3.5 \] Not divisible.
\[ 792 \div 132 = 6 \] Divisible.
\[ 968 \div 132 \neq \text{integer} \] Not divisible.
\[ 2178 \div 132 = 16.5 \] Not divisible.
\[ 5184 \div 132 \approx 39.27 \] Not divisible.
\[ 6336 \div 132 = 48 \] Divisible.
Thus, divisible numbers are: \[ 264,\ 396,\ 792,\ 6336 \] Total numbers divisible by \(132\): \[ 4 \]

Step 4: Final Answer:
The required count is: \[ \boxed{4} \] Hence, the correct option is: \[ \boxed{\text{(A) 4}} \]

Answer 2

To check whether each number is divisible by 132, it helps to break 132 into \(132 = 4 \times 3 \times 11\). A number is divisible by 132 only if it passes the divisibility tests for 4, 3, and 11 all at once. Let's check every number in the list this way.

1. 264: Last two digits 64 are divisible by 4. Digit sum \(2+6+4=12\) is divisible by 3. Alternating digit sum (right to left) \(4-6+2=0\) is divisible by 11. All three hold, so 264 is divisible by 132.
2. 396: Last two digits 96 are divisible by 4. Digit sum \(3+9+6=18\) is divisible by 3. Alternating sum \(6-9+3=0\) is divisible by 11. All three hold, so 396 is divisible by 132.
3. 462: Last two digits 62 are not divisible by 4 (\(62 \div 4 = 15.5\)). This already fails, so 462 is not divisible by 132.
4. 792: Last two digits 92 are divisible by 4. Digit sum \(7+9+2=18\) is divisible by 3. Alternating sum \(2-9+7=0\) is divisible by 11. All three hold, so 792 is divisible by 132.
5. 968: Last two digits 68 are divisible by 4, but the digit sum \(9+6+8=23\) is not divisible by 3. This fails, so 968 is not divisible by 132.
6. 2178: Last two digits 78 are not divisible by 4 (\(78 \div 4=19.5\)). This fails, so 2178 is not divisible by 132.
7. 5184: Last two digits 84 are divisible by 4, and the digit sum \(5+1+8+4=18\) is divisible by 3, but the alternating sum \(4-8+1-5=-8\) is not divisible by 11. This fails, so 5184 is not divisible by 132.
8. 6336: Last two digits 36 are divisible by 4. Digit sum \(6+3+3+6=18\) is divisible by 3. Alternating sum \(6-3+3-6=0\) is divisible by 11. All three hold, so 6336 is divisible by 132.

Only four numbers pass all three divisibility tests together: 264, 396, 792, and 6336.

Therefore, the correct answer is 4.