About fractions whose sum is a natural number Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each square with the numbers 1,2…30 without repeating any number, such that their sum is an integer number. square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square+square/square I got a solution by trial-and-error and it's: 14/1+23/2+11/22+19/3+10/15+29/4+9/12+17/6+5/30+25/8+21/24+16/7+20/28+27/18+13/26=53.I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.

Javion Henry 2022-07-17 Answered
About fractions whose sum is a natural number
Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each with the numbers 1 , 2 30 without repeating any number, such that their sum is an integer number.
+ + + + + + + + + + + + + +
I got a solution by trial-and-error and it's:
14 1 + 23 2 + 11 22 + 19 3 + 10 15 + 29 4 + 9 12 + 17 6 + 5 30 + 25 8 + 21 24 + 16 7 + 20 28 + 27 18 + 13 26 = 53.
I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.
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Answers (1)

Bianca Chung
Answered 2022-07-18 Author has 16 answers
Here's what one can play around with (and supposedly, Ian Miller started similarly):
We can build as many integer fractions as possible, namely
(1) 30 15 + 28 14 + 26 13 + 24 12 + 22 11 + 20 10 + 18 9 + 16 8 ,
each summand equalling 2
As 14 is already in use, we continue with
(2) 21 7 .
All multiples of 6 are used, so we postpone that for a moment and continue with
(3) 25 5 .
Now we are left with 29 , 27 , 23 , 19 , 17 , 6 , 4 , 3 , 2 , 1 to form five fractions. As barak manos observed, we must have the high primes 29 , 23 , 19 , 17 in numerators (or otherwise, 20 ! p times our sum would not be an integer - even if we had not started with the above strategy). Because of what we did so far, we also cannot have 27 in the denominator (or otherwise 12 times our sum would not be an integer) - but that might be different if we had started differently.
Now we try to exploit that 1 6 + 1 3 + 1 2 is an integer. We observe that 29 1 ( mod 6 ), 23 1 ( mod 3 ), and e.g. 27 1 ( mod 2 ), so that
(4) 29 6 + 23 3 + 27 2
is an integer.
We are now left with 19 , 17 , 4 , 1, and definitely end with something having a 4 in the denominator, namely
(5) 19 4 + 17 1 or 17 4 + 19 1 .
We play around with the previous fractions to mend this. For example, we can dissect our 28 14 and 21 7 to form 7 28 + 21 14 = 7 4 from its parts. This adjusts our quarter-integer result to either an integer result (and we are done) or to a half-integer result. Even though we can verify that we can guarantee an integer here by making the right choice in ( 5 ), it should be noted that by flipping one of the fractions in ( 1 ), we could turn a half-integer sum into an integer if necessary.

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