# Inverted numbers

talzamir
Posers and Puzzles 28 Jun '12 10:13
1. talzamir
Art, not a Toil
28 Jun '12 10:13
How many pairs of integers (x, y) are there that satisfy the equation

1 / x + 1 / y = 1 / 11 ?

Could a generic solution be found for a 1 / x + 1 / y = 1 / a where a is some integer > 0 ?
2. 29 Jun '12 11:58
Rearranging gives

y = 11*x/(x-11)

we need whole number values for y, most obvious is x = 12 which gives y = 132, there's also x = 22, y = 22.

For the general case, I guess for prime numbers the only values would be x = a + 1, y = (a+1)*a and x = 2*a, y = 2*a. Still working on the details of testing for things being coprime!
3. talzamir
Art, not a Toil
29 Jun '12 14:48
Any with negative values of x or y?
4. 29 Jun '12 15:39
Originally posted by talzamir
Any with negative values of x or y?
Good point, assuming that y = is negative (y = -z) we get

z = 11*x/(11-x)

which gives integer solutions for x = 10 corresponding to z = 110, can't see any other negative solutions
5. talzamir
Art, not a Toil
29 Jun '12 20:42
That would just about cover it, yes. ðŸ™‚

1/x + 1/y = 1/11
0 = xy - 11x - 11y
121 = xy - 11x - 11y + 121
121 = (x - 11) (y - 11)

121 can be a factor of two integers as
11 x 11 ; which gives 1 / 22 + 1 / 22 = 1 / 11
121 x 1 ; which gives 1 / 132 + 1 / 12 = 1 / 11
(-121) x (-1) ; which gives 1 / (-110) + 1 / 10 = 1 / 11
(-11) x (-11) ; which gives no solution as x and y can't be zero.

The same as you pointed out works for all primes. Composite numbers can be a bit trickier as there are so many ways to derive them by multiplying two integers together.
6. 30 Jun '12 14:323 edits
Originally posted by talzamir
That would just about cover it, yes. ðŸ™‚

1/x + 1/y = 1/11
0 = xy - 11x - 11y
121 = xy - 11x - 11y + 121
121 = (x - 11) (y - 11)

121 can be a factor of two integers as
11 x 11 ; which gives 1 / 22 + 1 / 22 = 1 / 11
121 x 1 ; which gives 1 / 132 + 1 / 12 = 1 / 11
(-121) x (-1) ; which gives 1 / (-110) + 1 / 10 = 1 / 11
(-11) x (-11) ; which gives e a bit trickier as there are so many ways to derive them by multiplying two integers together.
rearrange to y = 11x/(x-11)

if x is n*11 we have:
Y = 11*11*n/(n-1)*11
= 11n/(n-1)
n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:
x = 0
x = 22
x = -110
x = 132

let us try for solutions where x = n*11 + b, where b is between 1 and 10

then y = 11*(n*11 + b)/(n*11+b - 11)

Here the top is clearly a multiple of11, and the bottom is clearly NOT a multiple of 11, so this only has integer solutions when the bottom is plus or minus 1

so this gives us the solutions
n=0, b = 10, x = 10
n=1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132
7. SwissGambit
Caninus Interruptus
30 Jun '12 17:162 edits
Originally posted by iamatiger
rearrange to y = 11x/(x-11)

if x is n*11 we have:
Y = 11*11*n/(n-1)*11
= 11n/(n-1)
n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:
x = 0
x = 22
x = -110
x = 132

let us try for solutions where x ...[text shortened]... =1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132
Interesting that in the original equation, x=0 can't be a solution, but it can once you solve the equation for y. Once solved for y, x can't be 11. But x=11 was not going to produce a=11 in the original equation, so I guess we just manipulate things as needed to give the highest number of solutions. ðŸ˜›
8. 01 Jul '12 12:37
Originally posted by iamatiger
rearrange to y = 11x/(x-11)

if x is n*11 we have:
Y = 11*11*n/(n-1)*11
= 11n/(n-1)
n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:
x = 0
x = 22
x = -110
x = 132

let us try for solutions where x ...[text shortened]... =1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132
Yes 0 doesn't work as a solution because in the following:

1/x + 1/y = 1/11

to simplify this, we take 1/x from both sides, this doesn't work if x is 0 because then we are taking infinity from both sides, so we have assumed x /= 0, so we get to

1/y = 1/11 - 1/x {x =/ 0}

that assumption then carries through, so I should have got the pairs:

-1/110 + 1/10 = 1/11
1/12+ 1/132 = 1/11
1/22 + 1/22 = 1/11

which are the only possibilities
9. SwissGambit
Caninus Interruptus
01 Jul '12 19:014 edits
Originally posted by talzamir
Could a generic solution be found for a 1 / x + 1 / y = 1 / a where a is some integer > 0 ?
No one has posted the generic solution yet, so here it goes.

Solving the initial equation for y, we get:
y=ax/(x-a) [x!=0, y!=0]
to get a whole number for y, x = a+1 or a-1, or
x is some multiple of a, say ca. then
y=ca^2/(ca-a)
=ca^2/a(c-1)
=ca/(c-1)
...such that the fraction makes a whole number
obviously c=2 works, but not c=0 because that would make y=0
the only other way to get a whole number is to make the denominator a or -a. Making the denominator magnitude any bigger means that c/(c-1) is forever (in magnitude) approaching 1 without actually hitting it; thus not evenly divisible.
So,
c=2 (again) and c = 1 - a
...and that is all the solutions, for any integer a. [a!=0]

x=a+1 y=a^2+a
x=a-1 y=a-a^2
x=2a y=2a
x=a-a^2 y=a-1
10. 01 Jul '12 22:571 edit
Originally posted by SwissGambit
No one has posted the generic solution yet, so here it goes.

Solving the initial equation for y, we get:
y=ax/(x-a) [x!=0, y!=0]
to get a whole number for y, x = a+1 or a-1, or
x is some multiple of a, say ca. then
y=ca^2/(ca-a)
=ca^2/a(c-1)
=ca/(c-1)
...such that the fraction makes a whole number
obviousl ...[text shortened]...

x=a+1 y=a^2+a
x=a-1 y=a-a^2
x=2a y=2a
x=a-a^2 y=a-1
Generalising talzamir's approach

1/x + 1/y = 1/a
0 = xy - ax - ay
a^2 = xy - ax - ay + a^2
a^2 = (x - a) (y - a)

So take all the ways of factoring a^2 into two integers, and in each case add A to each factor to get x and y.

Lets try 1/x + 1/y = 1/6

6^2 is 36 = 2*2*3*3

So factorisations are:
1*36,2*18,3*12,4*9,6*6, plus all of those combos again with both numbers negated

Giving the combinations x&y of:
1/2-1/3
1/3-1/6
1/4-1/12
1/5-1/30
1/7+1/42
1/8+1/24
1/9+1/18
1/10+1/15
1/12+1/12
11. SwissGambit
Caninus Interruptus
02 Jul '12 00:282 edits
Originally posted by iamatiger
Generalising talzamir's approach

1/x + 1/y = 1/a
0 = xy - ax - ay
a^2 = xy - ax - ay + a^2
a^2 = (x - a) (y - a)

So take all the ways of factoring a^2 into two integers, and in each case add A to each factor to get x and y.

Lets try 1/x + 1/y = 1/6

6^2 is 36 = 2*2*3*3

So factorisations are:
1*36,2*18,3*12,4*9,6*6, plus all of those c ...[text shortened]... y of:
1/2-1/3
1/3-1/6
1/4-1/12
1/5-1/30
1/7+1/42
1/8+1/24
1/9+1/18
1/10+1/15
1/12+1/12
Yeah, I guess mine only works when a is a prime number.

Jeez, and I missed -110, 10. Really cocked this up. ðŸ˜ž
12. 02 Jul '12 07:34
And, there is no instantaneous way to find out the number of ways of factoring a number (otherwise there would be an instant way of finding prime numbers).
13. 02 Jul '12 07:41
Originally posted by SwissGambit
Yeah, I guess mine only works when a is a prime number.

Jeez, and I missed -110, 10. Really cocked this up. ðŸ˜ž
No, you got -110,10

that is your:
x=a-a^2 y= a-1

don't despair! ðŸ™‚
14. SwissGambit
Caninus Interruptus
02 Jul '12 18:20
Originally posted by iamatiger
No, you got -110,10

that is your:
x=a-a^2 y= a-1

don't despair! ðŸ™‚
Oh dear. I have to have missed something because I should have five solutions.

So which one did I not swap x and y for?

x=a^2+a y=a+1

x=132, y=12

Yeah, that's the one I missed. I guess I should have read your earlier post. You had five solutions. I should have known my generalization was wrong just from that.