Question

If n is positive even integer, then $n(n+1)(n+2)$ is ................

• A. a prime number
• B. divisible by 20
• C. divisible by 24
• D. divisible by 16

Answer 1

Answer: (C)

divisible by 24

Here $n$ is a positive even integer for some $m$ $\in N$ and $n=2m$.
Then $n(n+1)(n+2)=2m(2m+1)(2m+2)=4m(m+1)(2m+1)$ ........ $\left(1\right)$
Case 1. If m = 1
$m(m+1)(2m+1)=1(1+1)\left(2\right(1)+1)=2\times 3=6$ which is divisible by $6$.
$\therefore m(m+1)(2m+1)$ is divisible by $6$.
$\therefore 4m(m+1)(2m+1)$ is divisible by $24$.
Hence, $n(n+1)(n+2)$ is divisible by $24$. ......... (by $\left(1\right)$)
Case 2: If m = 2
$m(m+1)(2m+1)=2(2+1)\left(2\right(2)+1)=2\times 3\times 5=30$ which is divisible by $6$.
$\therefore m(m+1)(2m+1)$ is divisible by $6$.
$\therefore 4m(m+1)(2m+1)$ is divisible by $24$.
Case 3: If m $\ge$ 3
Here $m$ and $m+1$ being consecutive integers, one of them will always be even and the other will be odd.
$\therefore m(m+1)(2m+1)$ is always divisible by $2$
Also, $m(m\ge 3)$ is a positive integer, so for some $k\in N$, $m=3k$ or $m=3k+1$ or $m=3k+2$
i) For $m=3k$
$m(m+1)(2m+1)=3k(3k+1)\left(2\right(3k)+1)$
$=3\left[k\right(3k+1\left)\right(6k+1\left)\right]$
This is divisible by $3$.
ii) For $m=3k+1$
$m(m+1)(2m+1)=(3k+1)(3k+1+1)\left(2\right(3k+1)+1)$
$=\left[\right(3k+1\left)\right(3k+2\left)\right(6k+3\left)\right]$
$=3\left[\right(3k+1\left)\right(3k+2\left)\right(2k+1\left)\right]$
This is also divisible by $3$.
iii) For $m=3k+2$
$m(m+1)(2m+1)=(3k+2)(3k+2+1)\left(2\right(3k+2)+1)$
$=\left[\right(3k+2\left)\right(3k+3\left)\right(6k+5\left)\right]$
$=3\left[\right(k+1\left)\right(3k+2\left)\right(6k+5\left)\right]$
This is also divisible by $3$.
Hence, in any case $m(m+1)(2m+1)$ is divisible by $3$ and $2$.
As $2$ and $3$ are mutually prime numbers, $m(m+1)(2m+1)$ is divisible by $6$.
$\therefore n(n+1)(n+2)=4m(m+1)(2m+1)$ is divisible by $24$ ....... By $\left(1\right)$
Thus in any case $n(n+1)(n+2)$ is divisible by $24$

Hence, $n(n+1)(n+2)$ is divisible by $24$.