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

(A) \dfrac{2^{n}}{5^{n}}

(B) 4^{n}–\dfrac{2^{n}}{5^{n}}

(C) \dfrac{4^{n}}{5^{n}}

(D) none of these

**Answer **: option (A)

**Solution **:

In any number the last digit can be 0, 1, 2, 3,4, 5, 6, 7, 8, 9. Therefore, last digit of each number can be chosen in 10 ways. Thus, exhaustive number of ways is 10^{n}. If the last digit be 1, 3, 7 or 9, then none of the numbers can be even or end in 0 or 5.

Thus, we have a choice of 4 digits, viz, 1,3, 7 or 9 with which each of n numbers should end. So

favorable number of ways is 4^{n}. Hence, the required probability is

\dfrac{4^{n}}{10^{n}}=\left( \dfrac{4}{10}\right) ^{n}=\left( \dfrac{2}{5}\right) ^{n}=\dfrac{2^{n}}{5^{n}}

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