×

# Algebra Question

Find all the solutions of 1/x + 1/y = 1/2013.

I have found 8 solutions but one of my friend claims that it has 14 solutions. He told me that there is a trick for this.

Do any of you know the trick or help me find 14 solutions?

Thanks everyone!

Note by Anoopam Mishra
4 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

$$\frac{1}{x} + \frac{1}{y}= \frac{1}{2013} \implies xy= 2013(x+y)$$ This implies $$(x-2013)(y-2013) = xy - 2013(x+y) + 2013^2 = 2013^2$$.
Now note that $$2013^2= 3^2*11^2*61^2$$. So number of divisors (positive and negative both included) of $$2013^2$$ is $$2(2+1)(2+1)(2+1) = 54$$. For each divisor of $$2013^2$$ we get 1 possible value of $$((x-2013), (y-2013))$$. However note that $$(x, y)= (0, 0)$$ satisfies $$xy= 2013(x+y)$$, but not the original equation. So this case also has to be excluded. Thus from the total number of solutions we should subtract $$1$$. That would give the total number of solutions= $$54-1= 53$$.

- 4 years, 8 months ago

There are actually 53 distinct integer ordered pairs $$(x,y)$$ that satisfy the equation, not 52. You might want to double check your reasoning.

- 4 years, 8 months ago

Thank You!

- 4 years, 8 months ago