The product of a non-zero whole number and its successor is always
an even number
an odd number
a prime number
divisible by 3
Idea: Two numbers that come one after the other are called consecutive. Examples: (1, 2), (5, 6), (10, 11).
Let the whole number be \(n\). Because it is non-zero, \(n \ge 1\).
Its successor (the next number) is \(n + 1\).
Among any two consecutive numbers, one is always even.
Case 1: If \(n\) is even, then \(n = 2k\) for some whole number \(k\).
Case 2: If \(n\) is odd, then \(n = 2k + 1\). The next number is \(n+1 = 2k + 2 = 2(k+1)\), which is even.
So in both cases, at least one of \(n\) or \(n+1\) is a multiple of \(2\).
Therefore, their product has a factor \(2\): \(n(n+1)\) is even.
Quick check with examples:
\(3 \times 4 = 12\) (even), \(6 \times 7 = 42\) (even)
Conclusion: The product is always an even number (Option A).