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Let m=77^n . The index n is given a positive integral value at random. What is the probability that the value of m will have 1 in the units place?

Options

  1. A1 2
  2. B1 3
  3. C1 4
  4. D1 n

Correct answer

C. 1 4

Step-by-step solution

m=77^n aligned Put, n & =1,2,3,4 For, n & =1 m & =77 aligned For, n=2, m=77^2=5929 aligned For, n & =3, m=77^3=456543 For, n & =4, m=77^4=35153041 & = last digit will be 1 aligned Hence, the required probability = 1 4

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