How many ordered pairs of integers (a,b) are there such that 1/a+1/b=1/200?

Question

How many ordered pairs of integers (a,b) are there such that 1/a+1/b=1/200?

Answer 1

multiply both sides by 200ab

200b+200a = ab
200b - ab = -200a
b(200-a) = -200a
b = -200a/(200-a) = 200a/(a-200)
or
a = 200b/(b-200)

let a = 201, then b = 40200
let a = 199 , then b = -39800
let a = 202 , then b = 20200
let a = 198 , then b = -19800
let a = 203 , then b = not an integer !
let a = 197 , then b = not an integer!
let a = 204 , then b = 10200
let a = 205 , then b = 8200
let a = 206 , then b = not an integer, ahhh!
...
let a = 240, then b = 1200
let a = 600 , then b = 240000
let a = 10200 , then b = 204

we have to have (a-200) be divisible into 200a
obviously we cannot use a = 200, but we can go up or down from 200 leaving divisors of ±1 , ±2, ±4, ±5, ±8, ±10, .. as long as that divisor divides evenly into 200a
that is, 200 = 2x2x2x5x5

Wolfram says that there are 69 of these

http://www.wolframalpha.com/input/?i=%28x%2By%29%2F%28xy%29+%3D+1%2F200

Answer 2

This problem has been posted on Brilliant as a current problem in this weeks' Geometry problem set.

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Please refrain from telling the OP the answer.

I ask that moderators take this posting down.

- Calvin Lin
Brilliant Challenge Master

Answer 3

To find how many ordered pairs of integers (a, b) satisfy the equation 1/a + 1/b = 1/200, we can start by rearranging the equation.

Multiply both sides of the equation by 200ab to eliminate the denominators:

200b + 200a = ab

Now, we want to count the pairs of integer solutions. To simplify the equation, let's subtract 200a from both sides:

200b = ab - 200a

Next, we can factor the right side of the equation:

200b = a(b - 200)

Since the left side is divisible by 200, the right side must also be divisible by 200. Therefore, b - 200 must be divisible by 200.

Let's rewrite the equation in terms of divisibility:

b - 200 = 200k

where k is an integer.

Now, we can solve for b:

b = 200k + 200

This tells us that b is equal to 200 times an integer k plus 200.

With this information, we can conclude that b will be an integer if and only if k is an integer. So, b has infinite solutions.

For each value of b, we can find the corresponding a value using the originally rearranged equation:

200b + 200a = ab

Substituting the expression for b, we get:

200(a + k) = a(200k + 200)

Dividing both sides by 200, we get:

a + k = ak + 1

Rearranging this equation, we have:

a - ak = 1 - k

Factoring out 'a' from the left side, we get:

a(1 - k) = 1 - k

Now, if 1 - k = 0, we have no solution because it will lead to division by 0 in the original equation.

So, the valid case is when 1 - k ≠ 0.

Dividing both sides by 1 - k, we obtain:

a = (1 - k)/(1 - k)

a = 1

Therefore, for any value of b, the corresponding value of a is 1.

In conclusion, there are infinite ordered pairs (a, b) that satisfy the equation 1/a + 1/b = 1/200.