Term 2 Week 5 Junior Problems
Problem 1
Find a positive integer solution $(x,y,z,a,b)$ for which
\[\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.\]Is your answer the only solution? If so, show why.
Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning.
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The perfect number $28=1+2+4+7+14$ provides a solution: $$ \frac1{28}+\frac1{14}+\frac17+\frac14+\frac12=\frac{1+2+4+7+14}{28}=1\;. $$ If they’ve been doing unit (or ‘Egyptian’) fractions, I’d expect some to see that since $\frac16+\frac13=\frac12$, $$ \frac16+\frac16+\frac16+\frac16+\frac13=1 $$ is a solution, though not a much more interesting one than the trivial solution. The choice of letters might well suggest the solution $$ \frac16+\frac16+\frac16+\frac14+\frac14\;. $$ A little playing around would show that $\frac14+\frac15=\frac9{20}$, which differs from $\frac12$ by just $\frac1{20}$; that yields the solution $$ \frac1{20}+\frac15+\frac14+\frac14+\frac14\;. $$ If I were the teacher, I’d hope that some kids would realize that since the average of the fractions is $\frac15$, in any non-trivial solution at least one denominator must be less than $5$, and at least one must be greater than $5$. Say that $x\le y\le z\le a\le b$. Clearly $x\ge 2$, so let’s try $x=2$. Then we need to solve $$ \frac1y+\frac1z+\frac1a+\frac1b=\frac12\;. $$ Now $y\ge 3$. Suppose that $y=3$; then $$ \frac1z+\frac1a+\frac1b=\frac16\;. $$ Now $1,2,$ and $3$ all divide $36$, and $\frac16=\frac6{36}$, so we can write $$ \frac1{36}+\frac1{18}+\frac1{12}=\frac{1+2+3}{36}=\frac6{36}=\frac16\;, $$ and we get another ‘nice’ solution, $$ \frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;. $$Problem 2
Suppose a standard $8\times8$ chessboard has two diagonally opposite corners removed, leaving $62$ squares. Is it possible to place $31$ dominoes of size $2\times1$ so as to cover all of these squares?