Which one of the following is the greatest number by which the product of three consecutive even numbers would be exactly divisible?
[amp_mcq option1=”12″ option2=”24″ option3=”48″ option4=”64″ correct=”option3″]
This question was previously asked in
UPSC CAPF – 2020
Their product is $P = (2n)(2n+2)(2n+4) = 2n \cdot 2(n+1) \cdot 2(n+2) = 8n(n+1)(n+2)$.
The product of three consecutive integers $n(n+1)(n+2)$ is always divisible by $3! = 6$ (as one is divisible by 3 and at least one is divisible by 2).
Therefore, the product $P = 8 \times [n(n+1)(n+2)]$ is always divisible by $8 \times 6 = 48$.
To find the greatest number that *exactly* divides the product, we consider the smallest possible product, which occurs when $n=1$: $2 \times 4 \times 6 = 48$.
Any number that divides all such products must divide 48. The largest divisor of 48 is 48 itself. Since we’ve shown all products are divisible by 48, the greatest such number is 48.