According to question, ab is divisible by c.
Thus, either a is divisibe by c and/or b is divisible by c.
Considering the options:

A: a might or might not be divisible by c. It is not necessarily false.
B: same reasoning as A
C: Let c be a prime number, it can still divide ab
D: If c is odd; ab+c
=ck+c
=c(k+1)
Since, value of k is unknown, we cannot say that k+1 is even or odd. Thus, ab+c might be even or odd.
E: If c is even; ab+c
=ck+c
=c(k+1)
Since, c is even, the value c(k+1) is also even. Hence, ab+c is also even.
Thus, it is this option which is false.

Answer: E

Could you please explain it more clearly?

Official Explanation

If ab is divisible by c, then a might be divisible by c (e.g., if a = 8, b = 3, and c = 2). Choice (A) is possible and thus incorrect.

Additionally, this numeric example demonstrates that choice (C) is possible and thus incorrect, since 2 is a prime number.

If ab is divisible by c, then a might also not be divisible by c (for example, if a = 3, b = 4, and c = 6).Choice (B) is possible and thus incorrect.

In choice (D), if c is odd and ab + c is odd, then ab must be even. An even divided by an odd can result in an integer (e.g., if ab = 12 and c = 3). Choice (D) is possible and thus incorrect.

Finally, in choice (E), if c is even and ab + c is odd, then ab must be odd. An odd divided by an even will never result in an integer; that is, if ab is odd and c is even, ab will never be divisible by c.

Choice (E) is impossible and therefore is the correct answer.

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