How many 5 digit numbers are there, whose i digit is divisible by i?

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.


There are two mistakes:

  1. $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)
  2. There is no point in using permutations.

Tags:

Combinatorics

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