Question:

A, B, C, D, E and F are six whole numbers. Is "ABCDEF" divisible by 132?

Statement 1: The last four digits of the given number have a factor of 4 and \( A + C + E = 12(B + D + F) \)
Statement 2: The sum of all the digits of the given number is divisible by 24

Show Hint

Recall \( 132 = 4 \times 3 \times 11 \) and test each factor separately against both statements.
Updated On: Jul 21, 2026
  • If the data in statement (1) alone is sufficient to answer the question
  • If the data in statement (2) alone is sufficient to answer the question
  • If the data in both the statements together are needed to answer the question
  • If either statement (1) alone or statement (2) alone is sufficient to answer the question
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The Correct Option is C

Solution and Explanation

Step 1: Break 132 into its prime factors.
\( 132 = 4 \times 3 \times 11 \), so ABCDEF must pass the divisibility test for 4, for 3, and for 11 together.

Step 2: Test statement 1 alone.
The last four digits carrying a factor of 4 confirms the whole number is divisible by 4.
The 11-rule needs the alternating sum \( (A + C + E) - (B + D + F) \).
Using \( A + C + E = 12(B + D + F) \), this alternating sum becomes \( 12(B+D+F) - (B+D+F) = 11(B+D+F) \), always a multiple of 11.
So statement 1 secures divisibility by 4 and by 11.
But the total digit sum is \( 13(B+D+F) \), and this is a multiple of 3 only when \( (B+D+F) \) itself is a multiple of 3, which is not fixed.
Divisibility by 3 stays undecided, so statement 1 alone is not enough.

Step 3: Test statement 2 alone.
A digit sum divisible by 24 guarantees divisibility by 3, since 24 is a multiple of 3.
It says nothing about the last digits or the alternating sum, so divisibility by 4 and by 11 stay unknown.
Statement 2 alone is not enough either.

Step 4: Combine both statements.
Statement 1 secures divisibility by 4 and by 11.
Statement 2 secures divisibility by 3.
Together, ABCDEF is divisible by 4, by 3 and by 11, hence divisible by 132.

Final Answer:
Both statements are needed together to settle the divisibility. \[ \boxed{(c)} \]
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