Find the number of pairs of coprime integers (n,m) such that 1≤n≤m≤20.
To find the number of pairs of coprime integers (n,m) such that 1≤n≤m≤20, we need to count the number of pairs where n and m have no common factors other than 1.
One way to approach this problem is to use the principle of inclusion-exclusion. We can start by counting all possible pairs of integers (n,m) where 1≤n≤m≤20. There are a total of 20*21/2 pairs, which equals 210.
Next, we need to subtract the number of pairs where n and m have at least one common factor other than 1. To do this, we can iterate through each possible factor and count the number of pairs.
For example, let's consider the factor 2. We need to count the number of pairs (n,m) where both n and m are divisible by 2. Since 1≤n≤m≤20, n must be an even number between 2 and 20. There are a total of 10 even numbers in this range (2, 4, 6, ..., 20). For each even number n, the value of m can range from n to 20, inclusive. Therefore, there are 10 pairs for each factor.
Following the same logic, we can count the pairs for each factor from 2 to 20. Here is a table showing the number of pairs for each factor:
Factor | Number of Pairs
--------|----------------
2 | 10
3 | 6
5 | 4
7 | 2
11 | 1
13 | 1
17 | 1
19 | 1
Now, we need to subtract the total number of pairs for each factor from our initial count of all pairs. Starting with 210, we subtract the number of pairs for each factor:
210 - 10 - 6 - 4 - 2 - 1 - 1 - 1 - 1 = 184
Therefore, there are 184 pairs of coprime integers (n,m) such that 1≤n≤m≤20.