Question

plzz solve the problem in brief. ​

Answer 1

Let a and b be two consecutive positive integer.

Let b = 2

Therefore, possible remainders are 0 and 1.

According to Euclid's Division Lemma

a = bq + r, 0<or = r<b

a = 2q

a = 2q +1

Hence, the product of two consecutive positive integers is divisivle by 2.

Answer 2

ANSWER:

• PRODUCT OF TWO CONSECUTIVE INTEGER IS DIVISIBLE BY 2. (PROVED)

GIVEN:

• Two consecutive integers.

TO PROVE:

• The product of two consecutive integer is divisible by 2.

SOLUTION:

Let the first integer be 'n' then consecutive integer = (n+1)

Product of consecutive integer = n(n+1). ......(i)

when n is divided by 2 we get some quotient 'q' and some remainder 'r'.

n= 2q+r .....(ii) (standard form)

Case 1 :Putting r= 0 in eq (ii) we get;

n= 2q

putting n = 2q in eq (i) we get;

=2q(2q+1)

it is divisible by 2.

Case 2: Putting r= 1 in eq(ii) we get;

n= 2q+1

putting n= 2q+1 in eq (i) we get;

= (2q+1)(2q+1+1)

= 2(2q+1)(q+1). (Taking 2 as common)

It is also divisible by 2

So Case 1 and case 2 satisfied.

HENCE PROVED

NOTE:

• in this way we can prove that the product of 3 consecutive integer is divisible by 6