The product of two consecutive integers is divisible by

Asked by admin @ 11/08/2021 in Math viewed by 316 People

"The product of two consecutive positive integers is divisible by 2". Is this statement true or false? Give reason.

Answered by admin @ 11/08/2021

Answer:

True, because the product of any two consecutive numbers, say n(n+1) will always be even as one out of n or (n+1) must be even.

example: let the first integer be x

then the second integer shall be x+1

then their product be x(x+1) = x²+x

(i) If x is even

then x = 2k

∴ x²+x= (2k)²+2k

=4k²+2k

=2(2k²+k)

hence divisible by two.

(ii)Let x be odd.

∴ x= 2k+1

∴ x²+x = (2k+1)²+2k+1

=(2k)²+8k+1+2k+1

=4k²+10k+2

=2(2k²+5k+1)

hence divisible by two/.

since bothe of our conditions satisfy the statement, we can say that the product of two consecutive integers is divisible by 2

or you can take other answer

Let a, a + 1 be two consecutive positive integers. By Euclid’s division lemma, we have a = bq + r, where 0 ≤ r < b For b = 2 , we have a = 2q + r, where 0 ≤ r < 2 ...(i) Putting r = 0 in (i), we get a = 2q, which is divisible by 2. a + 1 = 2q + 1, which is not divisible by 2. Putting r = 1 in (i), we get a = 2q + 1, which is not divisible by 2. a + 1 = 2q + 2, which is divisible by 2. Thus for 0 ≤ r < 2, one out of every two consecutive integers is divisible by 2. Hence, The product of two consecutive positive integers is divisible by 2.

The product of two consecutive integers is divisible by

or you can take other answer

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