Question
On adding any three consecutive even numbers, the summation (result) would always be divisible by which of the following ?
This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is
6
To solve this problem, let's consider three consecutive even numbers and explore their properties.
- Let the first even number be \(2n\), where \(n\) is an integer.
- The next consecutive even number will be \(2n + 2\).
- The third consecutive even number will be \(2n + 4\).
- Now, we need to find the sum of these three numbers: \(2n + (2n + 2) + (2n + 4)\).
- Simplify the expression: \(= 2n + 2n + 2 + 2n + 4\). \(= 6n + 6\).
- Notice that we can factor out a 6 from the expression: \(= 6(n + 1)\).
- The expression \(6(n + 1)\) clearly shows that the sum is always divisible by 6.
Therefore, the summation of any three consecutive even numbers is always divisible by 6. Hence, the correct answer is 6.
Let's justify and rule out the other options provided:
- 5: There is no inherent factor of 5 in our expression \(6(n + 1)\).
- 7: Similarly, the expression does not include a factor of 7.
- 11: The number 11 is not a factor of 6 or of the number \(n + 1\) by default.
Thus, out of the given options, the number by which the sum is always divisible is 6.
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