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Posers and Puzzles

Posers and Puzzles

  1. Standard member 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. Standard member 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. Standard member talzamir
    Art, not a Toil
    29 Jun '12 20:42
    That would just about cover it, yes. 🙂

    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 14:32 / 3 edits
    Originally posted by talzamir
    That would just about cover it, yes. 🙂

    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. Standard member SwissGambit
    Caninus Interruptus
    30 Jun '12 17:16 / 2 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. 😛
  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. Standard member SwissGambit
    Caninus Interruptus
    01 Jul '12 19:01 / 4 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 22:57 / 1 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. Standard member SwissGambit
    Caninus Interruptus
    02 Jul '12 00:28 / 2 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. 😞
  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. 😞
    No, you got -110,10

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

    don't despair! 🙂
  14. Standard member 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! 🙂
    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.