Question

If the digits of the number 12345 are randomly rearranged, what is the probability that the new number is divisible by
(i) 3 (ii) 9

Answer 1

(i) For a number to be divisible by 3, the sum of digits has to be divisible by 3. Here, we can see that the sum of the five digits is 15. It will be 15 irrespective of the order in which these digits are written. Hence, it is a sure event and its probability will be 1.

(ii) For a number to be divisible by 9, the sum of digits has to be divisible by 9. Here, we can see that the sum of digits is 15, which is not a multiple of 9. So, irrespective of the order of the digits, the five digit number will not be a multiple of 9 (impossible event). Thus, the probability is 0.