Question

The product of two consecutive integers is always divisible by 2.
[a] True
[b] False.

Answer 1

Hint: Use the fact that if n is odd, then n+1 is even and if n is even, then n+1 is odd. Use the fact that two consecutive integers are of form {n,n+1}. Think what happens when we multiply an odd integer and an even integer. Think whether the result will be even or odd. Alternatively use the fact that ${}^{n+1}{C}_2$ is an integer. Alternatively, you can use Euclid's division lemma to prove the result.

Complete step-by-step answer:
Let the integers be n and n+1
We have either n is even, or n is odd
If n is even, we have n+1 is odd.
Now, we know that if c divides a, then c divides ab.
Since 2 divides n, we have 2 divides n(n+1).
Hence n(n+1) is divisible.
If n is odd:
Then we have n+1 is even.
Now since 2 divides n+1, hence 2 divides n(n+1).
Hence n(n+1) is even.
Hence in both the cases n(n+1) is even.
Hence n(n+1) is always even.
Hence the product of two consecutive integers is always even.
Hence the given statement is true.

Note: [1] Alternatively, we have
The number of ways in which 2 objects can be selected from n+1 objects is an integer.
Hence ${}^{n+1}{C}_2$ is an integer.
So let ${}^{n+1}{C}_2=k$
Hence, we have
 $\begin{array}{rl}& {\displaystyle \frac{(n+1)!}{2!(n-1)!}}=k\\ & \Rightarrow {\displaystyle \frac{(n+1)\left(n\right)(n-1)!}{2(n-1)!}}=k\\ & \Rightarrow n(n+1)=2k.\end{array}$
Hence the product of two consecutive integers is divisible by 2.
Hence the product of two consecutive integers is even.
[2] Alternatively, we have
If n is an integer then by Euclid's division lemma, we have
n = 2q+r, where q is an integer and r = 0 or 1
Hence any integer is of one of the form 2q, 2q+1.
If n = 2q, then we have
n+1 = 2q+1
Hence n(n+1) = 2(q(2q+1)) which is even
If n = 2q+1, then we have
n+1 = 2q+2 = 2(q+1)
Hence n(n+1) = 2((2q+1)(q+1)), which is even.
Hence n(n+1) is always even.
Hence the product of two consecutive integers is always divisible by 2.
[3] product of r consecutive integers is divisible r!