“The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

(i) False As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6} ⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}  (ii) False As a ∈ {a, e, i,...

Solution: True. Justification: 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...

Solution: Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer. By Euclid’s division lemma, we have a = bq + r; 0 ≤...

Let the three consecutive integers bex, x+1and x+2. According to the question, x+x+1+x+2 = 51 => 3x+3=51 => 3x+3-3=51-3 [Subtracting 3 from both sides] => 3x=48 => 3x/3 = 48/3 [Dividing both sides by 3] X=16 Hence, first...

Let, (n – 1) and n be two consecutive positive integers ∴ Their product = n(n – 1) = n 2 −n We know that any positive integer is of the form 2q...

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q +...

True,because n(n+1) will always be even, as one out of the n or (n+1) must be even.

True, because n(n+1)(n+2) will always be divisible by 6, as least one of the factors will be divisible by 2 and at one of the factors will be divisible by...

(i) False As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6} ⇒ {2, 3, 4, 5} ∩ {3, 6} = {3} (ii) False As a ∈ {a, e, i, o, u}, a...

True, because n(n + 1) will always be even, as one out of the n or n+1 must be even.