Question

The sum of three consecutive odd numbers is always divisible by
A) 1
B) 3
C) 5
D) 6

Answer 1

Hint: The consecutive odd numbers refer to the numbers taken in a continuous manner skipping events in between. We can assume 3 such numbers using a variable, take their sum and see what number can we get common because that will be the one by which the sum will always be divisible

Complete step-by-step answer:
Some consecutive odd numbers are:
1, 3, 5, 7
We can see that the gap between two consecutive odd numbers is 2. Let us assume three consecutive numbers to be:
 $
  n \\
  n + 2 \\
  n + 2 + 2 \\
  \Rightarrow\left( {n,n + 2,n + 4} \right) \;
  $
The sum of these three odd consecutive numbers is given as:
 $ n + n + 2 + n + 4 = 3n + 6 $
From this sum, we can take number 3 as common:
 $ \left( {3n + 6} \right)\Rightarrow 3\left( {n + 2} \right) $
This shows that the answer will be a product of 3, n be any number. Thus, the same will be divisible by 3.
Therefore, the sum of three consecutive odd numbers is always divisible by 3 and the correct answer is $ 3 $ .
So, the correct answer is “Option B”.

Note: The natural numbers (greater than 1) are odd and even in an alternative manner and thus consecutive events and odds respectively are at a gap of two.
We can check if the answer obtained is correct or not by taking an example.
If the three consecutive integers are 3, 5, 7, their sum will be:
 $ 3 + 5 + 7 = 15 $
And 15 is the product of 3 and 5 and thus is divisible by both.
If the three consecutive integers are 11, 13, 15, their sum will be:
 $ 11 + 13 + 15 = 39 $
And 39 is the product of 3 and 11 and thus is divisible by both.
The common number by which both the sums are divisible is 3 so the answer we obtained is correct.