Question

The largest natural number by which the product of three consecutive even natural numbers is always divisible is?
(a) 6
(b) 24
(c) 48
(d) 96

Answer 1

Hint:Assume the three consecutive even natural numbers as 2n, 2n+2, 2n+4. Here, n is any odd number. Now, find all the possible factors of the product of these three consecutive even natural numbers. Multiply all the obtained factors to get the answer.

Complete step-by-step answer:
Let us assume the three consecutive natural numbers are of the form ‘2n’, ‘2n+2’, ‘2n+4’. 2 is multiplied to every number because we are assuming that ‘n’ is an odd number.
Now, taking the product of these numbers, we get,
2n(2n+2)(2n+4) = 8n(n+1)(n+2)
Clearly, we can see that, 8 is a factor of the given product.
Now, since n is an odd number, therefore (n+1) must be an even number and it must be divisible by at least 2.
Also, one of these numbers, n, (n+1), (n+2) must be a multiple of 3 because if we will choose ‘n’ as any non-multiple of 3 then either of the numbers (n+1) or (n+2) will be divisible by 3, which will depend upon the selection of ‘n’.
Therefore, we obtained three factors which are 2, 3 and 8. Multiplying these factors we get,
$\begin{align}
  & 2\times 3\times 8 \\
 & =48 \\
\end{align}$
Therefore, the largest natural number by which the product of three consecutive even natural numbers is always divisible is 48.
Hence option (c) is the correct answer.

Note: One may note that an alternate method or we can say hit and trial method can be, assuming the numbers as 2, 4 and 6. These are the smallest three consecutive even natural numbers. Multiplying these numbers, we get 48. Therefore, it is a proof that the largest natural number by which the product of three consecutive even natural numbers is always divisible is 48.