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

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:  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{align} & \dfrac{\left( n+1 \right)!}{2!\left( n-1 \right)!}=k \\ & \Rightarrow \dfrac{\left( n+1 \right)\left( n \right)\left( n-1 \right)!}{2\left( n-1 \right)!}=k \\ & \Rightarrow n\left( n+1 \right)=2k. \\ \end{align}
Hence the product of two consecutive integers is divisible by 2.
Hence the product of two consecutive integers is even.
 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.
 product of r consecutive integers is divisible r!

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