Question

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

Answer 1

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.

Answer 2

Answer:

THIS STATEMENT IS 100 PERCENT TRUE THAT PRODUCT OF TWO CONSECUTIVE POSITIVE INTEGERS S ALWAYS DIVISIBLE BY 2

THE PRODUCT OF TWO CONSECUTIVE NUMBERS .

i.e. n×(n+1)= will always be even and thus divisible by 2.

FOR e.g. 6×7= 42/2 =21

    9×10=90/2=45