(a) (a2 – b2) is divisible by 3

Question

The integers a, b and c are selected from 3n consecutive integers {1, 2, 3, …, 3n}. Then in how many ways can these integers be selected such that,

1. (a2 – b2) is divisible by 3

(A) n2 + nC2

(B) n2 + n + 1C2

(C) n2 + 3n C2

(D) of these

2. (a3 + b3) is divisible by 3

(a) 3n2 - n/2

(b) 3n2 + n/2

(c) n(n + 1)/2

(d) of these

3. Their sum is divisible by 3

(a) n/2(3n2 - 3n + 2)

(b) 3n2 – 3n + 2

(c) n/2

(d) of these

Answer 1

Correct option 1. (c) 2. (a) 3. (a)

1.a2 – b2 is divisible by 3, if either a + b, is divisible by 3 or a – b is divisible by 3 or both

G1 → a + b is 3λ type

G2 → a + b is 3λ - 1 type

G3 → a + b is 3λ - 2 type

Clearly, this is possible if either a and b are chosen from same group or one of them is chosen from G2 and other from G3 .

Therefore, the number of ways = nC2 + nC2 + nC2 + nC1.nC1

= 3n(n - 1)/2 + n2