**PROBLEM LINK**:

Practice

Author: Rupanjan Hari

Tester: Rupanjan Hari

**DIFFICULTY**

: Easy

**PREREQUISITES**

: Math

**QUICK EXPLANATION**

Given an integer **N**, traverse its digits (d**1**,d**2**,…,d**n**) and determine how many digits evenly divide **N**(i.e.: count the number of times **N** divided by each digit d**i** 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).

**EXPLANATION**

To make the problem look easy, I have divided the program into functions of appropriate name.

## Arrays:

a[] - The primary array that receives all the numbers## Functions:

int count(long long int n) - This function counts the digits that evenly divide the number.

In the main function just call the count function for each number.

**SOLUTION:**

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