If four whole numbers taken art random are multiplied together, then find the probability that the last digit in the product is 1,3,7, or 9.

Explanation

Let us call the numbers a, b, c, d and their product x=abcd. The digits at units place in each of the numbers a, b, c, d can be any one of the ten digits 0, 1, 2,…..9.
Thus the number of ways in which the digits at units place can be filled `=10xx10xx10xx10=10^4`.
Now the digit at unit place in x, i.e. the product abcd, will be 1 or 3 or 7 or 9 only when the digit at unit place in none of the no. a, b, c, d is 0, 2, 4, 6, 8, or 5.
Consequently the unit place in each of the numbers a, b, c, d can only be 1, 3, 7 or 9 if the product is to end in 1, 3, 7 or 9, so the no. of ways `=4xx4xx4xx4`.
Required probability `=4^4/10^4=(4/10)^4=(2/5)^4=16/625`