Four positive numerator-1 fractions summing to 3/7This year's seventh-grade olympiad sponsored by Tel Aviv University, round two, held a couple of days ago, had this (translated by yours truly) as its third question:Find four distinct natural numbers, a, b, c, and d, such that 1 a + 1 b + 1 c + 1 d = 3 7 My solution: 3 7 = 1 7 + . 285714 ¯ = 1 7 + 1 5 + .0 857142 ¯ = 1 7 + 1 5 + 6 70 = 1 7 + 1 5 + 1 70 + 1 14 Tweaking that a bit led me to 1 7 + 1 4 + 1 32 + 1 224 Question: Are there more solutions? Especially: Are there infinitely many?

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Four positive numerator-1 fractions summing to 3/7This year&#039;s seventh-grade olympiad sponsored by Tel Aviv University, round two, held a couple of days ago, had this (translated by yours truly) as its third question:Find four distinct natural numbers,   a,   b,   c, and   d, such that      1    a    +      1    b    +      1    c    +      1    d    =      3    7  My solution:                              3          7                            =                  1          7                +        .                  285714          ¯                                          =                  1          7                +                  1          5                +        .0                  857142          ¯                                          =                  1          7                +                  1          5                +                  6          70                                          =                  1          7                +                  1          5                +                  1          70                +                  1          14                    Tweaking that a bit led me to      1    7    +      1    4    +      1    32    +      1    224  Question: Are there more solutions? Especially: Are there infinitely many?

hyttnp · Accepted Answer

The greedy algorithm gives the triple       1    3    +      1    11    +      1    231  This leads to three quadruples using      1    a    =      1          a      +      1        +      1                  a        2            +      a      These would be denominator quadruples  (  4  ,  12  ,  11  ,  231  )  ↦  (  4  ,  11  ,  12  ,  231  )  ,  (  3  ,  12  ,  132  ,  231  )  ,  (  3  ,  11  ,  232  ,  53292  )  .There are finitely many. We demand that denominators be written in increasing order. The first can be no larger then   4  ⋅  (  7      /    3  )  . If   10 or larger,  4  (  1      /    10  )  =  2      /    5  &amp;lt;  3      /    7For each choice of the first denominator, there is likewise a bound on the second denominator, and so on.

Four positive numerator-1 fractions summing to 3/7 This year's seventh-grade olympiad sponsored by Tel Aviv University, round two, held a couple of days ago, had this (translated by yours truly) as its third question: Find four distinct natural numbers, a, b, c, and d, such that 1/a+1/b+1/c+1/d=3/7

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