(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 + 1}C₂

(C)  n² + 3ⁿ C₂ 

(D)  None of these

2.  (a³ + b³) is divisible by 3

(a)  3n² - n/2 

(b)  3n² + n/2

(c)  n(n + 1)/2

(d)  None of these

3.  Their sum is divisible by 3

(a)  n/2(3n² - 3n + 2)

(b)  3n²  – 3n + 2

(c)  n/2

(d)  None 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