Question

Sum of three consecutive odd numbers is always divisible by: 
A) 2 
B) 3 
C) 5 
D) 6 
 

• A. A and B only
• B. B and D only
• C. B only
• D. A and C only

Hint:

Whenever dealing with consecutive numbers, express them algebraically and factorize the sum to check divisibility conditions easily.

Answer 1

The Correct Option is C

Step 1: Assume three consecutive odd numbers.
Let the three consecutive odd numbers be:
\[ (2n+1), (2n+3), (2n+5) \]
Step 2: Find their sum.
\[ (2n+1) + (2n+3) + (2n+5) \]
\[ = 6n + 9 \]
Step 3: Factorize the sum.
\[ 6n + 9 = 3(2n + 3) \]
Step 4: Check divisibility by given numbers.
Divisibility by 2:
The expression \(3(2n+3)\) is not always even.
Hence, it is not always divisible by 2.
Divisibility by 3:
Since the sum is a multiple of 3, it is always divisible by 3.
Divisibility by 5:
The expression does not always contain factor 5.
Hence, it is not always divisible by 5.
Divisibility by 6:
To be divisible by 6, the number must be divisible by both 2 and 3.
Since it is not always divisible by 2, it is not divisible by 6.
Step 5: Final evaluation.
Only statement B is always true.
Step 6: Final conclusion.
Hence, the correct answer is B only.