How many pairs of positive integers (a,b), where a≤b satisfy 1/a+1/b=1/50?
well, you know that (100,100) is one pair. So, there are a maximum of 99 other pairs that will fit. That's a lot to check by hand, so there must be some theoretical way to resolve them.
1/4 + 1/4 = 1/2
1/6 + 1/3 = 1/2
1/6 + 1/6 = 1/3
1/12 + 1/4 = 1/3
1/8 + 1/8 = 1/4
1/12 + 1/6 = 1/4
1/20 + 1/5 = 1/4
Still looks tedious to run down the list starting at 1/100
1/b = 1/50 - 1/a = (a-50)/50a
b = 50a/(a-50)
If a = 100, b = 5000/50 = 100
If a = 75, b = 150
There are no other integer a values ¡Üb that yield an integer for b.
50/(a-50) would have to be an integer
well, what the correct answer?
To find the number of pairs of positive integers (a, b) that satisfy the equation 1/a + 1/b = 1/50 with a ≤ b, we can use algebraic manipulation and counting techniques.
1. Begin by multiplying both sides of the equation by 50ab to eliminate the denominators:
50b + 50a = ab
2. Rearrange the equation to obtain a quadratic equation in terms of a:
ab - 50a - 50b = 0
3. To solve for a, we can use the quadratic formula:
a = (50 ± √(2500 + 200b)) / 2
Since a and b are positive integers, the discriminant, 2500 + 200b, must be a perfect square.
4. Let's find the possible values of b:
- First, observe that the discriminant is always greater than or equal to 2500.
- We can set an upper bound on b by noting that a is positive, so:
(50 + √(2500 + 200b)) / 2 > 0
√(2500 + 200b) > -50
2500 + 200b > 2500
b > 0
Therefore, b can take values from 1 to infinity.
5. Check each value of b from 1 to infinity and see if 2500 + 200b is a perfect square.
- Calculate √(2500 + 200b) for each value of b.
- If √(2500 + 200b) is an integer, it means that b is valid.
- For each valid value of b, calculate the corresponding value(s) of a.
6. Count the number of valid pairs of (a, b) that satisfy the equation.
This method allows us to systematically find the pairs of positive integers (a, b) that satisfy the given equation.