# Test of mathematical intuition?

How many pairs of numbers x and y would you guess that there are there such that 1/x + 1/y = 1/10?

I will show a way of finding them all.

Observing members:
0
Composing members:
0
## 7 Answers

Deleted by me. I have to rethink it.

1+x/y=x/10

y+x=xy/10

10y+10x=xy

Looking at this, and I could be wrong, I would say that the only solution is if x and y both equal 0.

My intuition says not to care.

My mathematical intuition plus grade school line equation knowledge, says that’s a curve equation, and that there would be an infinite number of point solutions x & y, which also passes the test of imagining plugging in various numbers for x & y.

The first thing to realize is that x and y both have to be greater than 10.

Next, notice that half of 1/10 is 1/(20). One of 1/x and 1/y is greaer than or equal to 1/(20) and the other is less than or equal to 1/(20). We can arbitrarily choose 1/x >= 1/(20). That means that x<=20. Therefore possible x values go from 11 to 20, meaning there are at most 10 possible pairs. Using brute force we can check which values of x give us integer values for y. We get the following pairs:

1/11, 1/110

1/12, 1/60

1/14, 1(/35)

1/15, 1/(30)

1/(20), 1/(20)

I’m confused by your answer.

First, what are the parentheses doing?

Second, where did you constrain answers to integers? Seems to me you wrote numbers, and numbers include non-integers.

So how about x = 25, y = 16 + ⅔?

If you don’t constrain to integers, it seems to me there are an infinite number of pairs of values for x and y (the reciprocals of all pairs of numbers that add to 1/10). I don’t think x or y are bounded, either, since for any value x, y is the number such that their reciprocals add to 1/10.

I should have specified that x and y must be integers. Otherwise, there are infinitely many possibilities.

As for the parentheses, they were needed in order to have the numbers displayed prpoperly.

## Answer this question