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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 pn denote the probability that a2 – 1 is divisible by 10.

If r = 0, then pn equals 

(A) 2k/n 

(B) (k + 1)/n 

(C) (2k + 1)/n 

(D) k/n

1 Answer

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Best answer

Answer is (A) 2k/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.

n = 10k + r

r = 0

n = 10k, no. of a whose unit place is 1 or 9

⇒ k = 1, n = 10, no. of a whose unit place is 2

⇒ k = 2, n = 20, no. of a whose unit place is 4

⇒ k = k, n = 10k, no. of a whose unit place is 2k 

Therefore

pn = 2k/n

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