Question

The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Answer 1

Hint: In the given question, we have been given a statement. We have to prove it. To prove a statement, we consider all of its cases. Then we prove them sequentially (individually or together if some cases overlap) so that the given statement does not have any left-out piece. For a general or even a specific formula whose range is such that proving case by case is not possible, we use the theory of mathematical induction. It is a standard practice for proving the cases of any formula, statement, et cetera, whose possible values are infinite or very huge.

Complete step by step solution:
In the given question, we have to prove that the product of any three consecutive numbers is divisible by \[6\]. If a number is divisible by \[6\], then it means that it is also divisible by \[2\] and \[3\].
So, let us prove that the product of any three consecutive numbers is divisible by \[2\] and \[3\].
Consider the three consecutive numbers to be \[x,\left( {x + 1} \right),\left( {x + 2} \right)\].
For \[2\]:
If \[x\] is not divisible by \[2\], then it means that \[x\] is odd.
Hence, if \[x\] is odd, then it is a known fact that \[x + 1\] (any odd number plus \[1\]) is even, hence, is divisible by \[2\].
Thus, out of the three consecutive numbers, at least one of them is always divisible by \[2\].
For \[3\]:
Now, if a number is not divisible by \[3\] and is divided by it, then it can leave either of only two remainder – \[1\] and \[2\].
If \[x\] is not divisible by \[3\] and when divided by \[3\] leaves a remainder of \[1\], then \[x + 2\] is going to be divisible by \[3\].
If \[x\] is not divisible by \[3\] and when divided by \[3\] leaves a remainder of \[2\], then \[x + 1\] is going to be divisible by \[3\].
Similarly, for the other two cases – \[\left( {x + 1} \right)\] and \[\left( {x + 2} \right)\] not being divisible by \[3\], we can have that either of the other two are going to be divisible by \[3\].
Thus, out of the three consecutive numbers, exactly one of them is always divisible by \[3\].
Now, we have shown that out of the three consecutive numbers, one of them is always divisible by \[2\] and \[3\]. Hence, we showed that the product of any three consecutive numbers is always divisible by \[6\].
Now, let us show that using some examples:
Let the triplet be \[13,14,15\].
Product of the three numbers is \[13 \times 14 \times 15 = 2730\].
Now, \[2730 \div 6 = 455\]
Hence, the product of the given triplet is divisible by \[6\].
Let the triplet be \[16,17,18\].
Product of the three numbers is \[16 \times 17 \times 18 = 4896\].
Now, \[4896 \div 6 = 816\]
Hence, the product of the given triplet is divisible by \[6\].

Note: In the given question, we had to prove a given statement. In general, if we have to prove any statement, whose domain (the possible values which can be put as input) is small, we prove the statement sequentially so that we cover all the possible bases; the sequential proving can be individual or collective depending on the overlapping of the cases. If we have such a statement that we cannot prove case by case because of the large (or even infinite) number of possible values or domains, we use the theory of mathematical induction. In this question, the number of cases were only two, so we used to solve it case by case only.