#PROBLEM LINK: Contest Page | CodeChef
Author: Setter’s name
DIFFICULTY : EASY
#PREREQUISITES : NIL
Given an integer,N , traverse its digits (d1,d2,…,dn) and determine how many digits evenly divide N (i.e.: count the number of times N divided by each digit di has a remainder of 0 ). Print the number of evenly divisible digits.
Note: Each digit is considered to be unique, so each occurrence of the same evenly divisible digit should be counted (i.e.: for N=111, the answer is 3).
The number 12 is broken into two digits 1 and 2. When 12 is divided by either of those digits, the calculation’s remainder is 0; thus, the number of evenly-divisible digits in 12 is 2.
The number 1012 is broken into four digits 1, 2, 3 and 2. 1012 is evenly divisible by its digits 1, 1, and 2, but it is not divisible by 0 as division by zero is undefined; thus, our count of evenly divisible digits is 3.