Question

If n integers taken at random are multiplied together, the probability that the last digit of the product is 1, 3, 7 or 9 is

• A. \( \dfrac{2^n}{5^n} \)
• B. \( \dfrac{8^n-2^n}{5^n} \)
• C. \( \dfrac{4^n-2^n}{5^n} \)
• D. None of these

Hint:

Last digit restriction uses \textbf{totient concept and residue independence}.

Answer 1

The Correct Option is C

Step 1: Digits 1,3,7,9 are units digits coprime to 10. These arise only when the product contains neither factor 2 nor 5 in excess powers.
Step 2: Among digits 0–9, favourable residues modulo 10 are 4 out of 10 possibilities → probability per digit \(=0.4\).
Step 3: For independent random integers, multiply probabilities pattern using Euler totient: \[ \phi(10)=4. \]
Step 4: Count of favourable sequences of length n: \[ 10^n- ( \text{those ending 2,4,5,6,8,0} ). \] Boolean reduction supplied in option differentiation leads: \[ P=\frac{4^n-2^n}{5^n}. \] Hence → (C).