Question:hard

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
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: List what 132 demands.
Since \( 132 = 4 \times 3 \times 11 \), three separate checks must all pass: divisible by 4, divisible by 3, divisible by 11.

Step 2: Check statement 2 first.
A digit sum divisible by 24 covers the divisible-by-3 check, because 24 is itself a multiple of 3.
It gives no information about the last digits, so the divisible-by-4 check is open.
It also gives no information about the alternating digit sum, so the divisible-by-11 check is open.
Two of the three checks remain unresolved, so statement 2 by itself cannot decide the question.

Step 3: Check statement 1 next.
The last four digits carrying a factor of 4 settles the divisible-by-4 check.
For the divisible-by-11 check, form the alternating sum \( (A+C+E) - (B+D+F) \).
Replacing \( A+C+E \) with \( 12(B+D+F) \) turns this into \( 11(B+D+F) \), which is a multiple of 11 no matter what \( (B+D+F) \) equals.
So the divisible-by-11 check also passes.
Only the divisible-by-3 check is left open, since the total digit sum \( 13(B+D+F) \) is a multiple of 3 only for particular values of \( (B+D+F) \).
One of the three checks stays open, so statement 1 alone also falls short.

Step 4: Merge the two statements.
Statement 1 closes the divisible-by-4 and divisible-by-11 checks.
Statement 2 closes the divisible-by-3 check.
All three checks pass only when both statements are used together.

Final Answer:
The two statements combined confirm ABCDEF is divisible by 132. \[ \boxed{(c)} \]
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