Formatted question description: https://leetcode.ca/all/2176.html

# 2176. Count Equal and Divisible Pairs in an Array (Easy)

Given a **0-indexed** integer array `nums`

of length `n`

and an integer `k`

, return *the number of pairs*

`(i, j)`

*where*

`0 <= i < j < n`

, *such that*

`nums[i] == nums[j]`

*and*

`(i * j)`

*is divisible by*

`k`

.

**Example 1:**

Input:nums = [3,1,2,2,2,1,3], k = 2Output:4Explanation:There are 4 pairs that meet all the requirements: - nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2. - nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2. - nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2. - nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

**Example 2:**

Input:nums = [1,2,3,4], k = 1Output:0Explanation:Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

**Constraints:**

`1 <= nums.length <= 100`

`1 <= nums[i], k <= 100`

**Similar Questions**:

## Solution 1. Brute Force

```
// OJ: https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
int countPairs(vector<int>& A, int k) {
int N = A.size(), ans = 0;
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
ans += A[i] == A[j] && i * j % k == 0;
}
}
return ans;
}
};
```