If 1 ≤ r ≤ 8, then pn equals (A) (2k – 1)/n (B) (2k/n)

Question

Let n = 10k + r when k, r ∈ N, 0 ≤ r ≤ 9. A number a is chosen at random from the set {1, 2, ...., n} and let p_n denote the probability that a² – 1 is divisible by 10.

If 1 ≤ r ≤ 8, then p_n equals

(A) (2k – 1)/n 

(B) (2k/n) 

(C) (2k + 1)/n 

(D) k/n

Answer 1

Answer is (C) (2k + 1)/n

n = 10k + r, k, r ∈ N, 0 ≤ r ≤ 9 

Unit place of a² will contain 0, 1, 4, 5, 6, 9 only. 

Hence, a² – 1 is divisible by 10 only if unit place of a² contain 1.

If unit place of a² is 1, then unit place of a will be 1 or 9.

Number of a whose unit place is 1 or 9 = 2k + 1.

Therefore,

p_n = (2k + 1)/n