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. \[ \boxed{(c)} \]