Javion Henry

2022-07-17

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 $\overline{){\displaystyle}}$ with the numbers $1,2\dots 30$ without repeating any number, such that their sum is an integer number.

$\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}$

I got a solution by trial-and-error and it's:

$\frac{14}{1}+\frac{23}{2}+\frac{11}{22}+\frac{19}{3}+\frac{10}{15}+\frac{29}{4}+\frac{9}{12}+\frac{17}{6}+\frac{5}{30}+\frac{25}{8}+\frac{21}{24}+\frac{16}{7}+\frac{20}{28}+\frac{27}{18}+\frac{13}{26}=53.$

I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.

Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each $\overline{){\displaystyle}}$ with the numbers $1,2\dots 30$ without repeating any number, such that their sum is an integer number.

$\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}+\frac{\overline{){\displaystyle}}}{\overline{){\displaystyle}}}$

I got a solution by trial-and-error and it's:

$\frac{14}{1}+\frac{23}{2}+\frac{11}{22}+\frac{19}{3}+\frac{10}{15}+\frac{29}{4}+\frac{9}{12}+\frac{17}{6}+\frac{5}{30}+\frac{25}{8}+\frac{21}{24}+\frac{16}{7}+\frac{20}{28}+\frac{27}{18}+\frac{13}{26}=53.$

I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.

Bianca Chung

Beginner2022-07-18Added 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

$\begin{array}{}\text{(1)}& \frac{30}{15}+\frac{28}{14}+\frac{26}{13}+\frac{24}{12}+\frac{22}{11}+\frac{20}{10}+\frac{18}{9}+\frac{16}{8},\end{array}$

each summand equalling $2$

As $14$ is already in use, we continue with

$\begin{array}{}\text{(2)}& \frac{21}{7}.\end{array}$

All multiples of $6$ are used, so we postpone that for a moment and continue with

$\begin{array}{}\text{(3)}& \frac{25}{5}.\end{array}$

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, $\frac{20!}{p}$ 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 $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}$ is an integer. We observe that $29\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}6)$, $23\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)$, and e.g. $27\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2)$, so that

$\begin{array}{}\text{(4)}& \frac{29}{6}+\frac{23}{3}+\frac{27}{2}\end{array}$

is an integer.

We are now left with $19,17,4,1$, and definitely end with something having a $4$ in the denominator, namely

$\begin{array}{}\text{(5)}& \frac{19}{4}+\frac{17}{1}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{17}{4}+\frac{19}{1}.\end{array}$

We play around with the previous fractions to mend this. For example, we can dissect our $\frac{28}{14}$ and $\frac{21}{7}$ to form $\frac{7}{28}+\frac{21}{14}=\frac{7}{4}$ 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.

We can build as many integer fractions as possible, namely

$\begin{array}{}\text{(1)}& \frac{30}{15}+\frac{28}{14}+\frac{26}{13}+\frac{24}{12}+\frac{22}{11}+\frac{20}{10}+\frac{18}{9}+\frac{16}{8},\end{array}$

each summand equalling $2$

As $14$ is already in use, we continue with

$\begin{array}{}\text{(2)}& \frac{21}{7}.\end{array}$

All multiples of $6$ are used, so we postpone that for a moment and continue with

$\begin{array}{}\text{(3)}& \frac{25}{5}.\end{array}$

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, $\frac{20!}{p}$ 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 $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}$ is an integer. We observe that $29\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}6)$, $23\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)$, and e.g. $27\equiv -1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2)$, so that

$\begin{array}{}\text{(4)}& \frac{29}{6}+\frac{23}{3}+\frac{27}{2}\end{array}$

is an integer.

We are now left with $19,17,4,1$, and definitely end with something having a $4$ in the denominator, namely

$\begin{array}{}\text{(5)}& \frac{19}{4}+\frac{17}{1}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{17}{4}+\frac{19}{1}.\end{array}$

We play around with the previous fractions to mend this. For example, we can dissect our $\frac{28}{14}$ and $\frac{21}{7}$ to form $\frac{7}{28}+\frac{21}{14}=\frac{7}{4}$ 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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