# 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

Deleted by me. I have to rethink it.

JLeslie (65413)

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.

Zaku (30353)

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)

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.

Zaku (30353)

Here is a graph, FWIW.

Zaku (30353)

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.