SOLUTION TO PROBLEM H-7

If $k \neq 1$ is an integer, then $\frac{1}{k}<\frac{1}{2}$. (k is nonzero of course). It follows that none of the numbers x, y, z can be negative. Indeed, if we assume for example that x < 0 then

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<0+\frac{1}{2}+\frac{1}{2}=1$$

Thus x, y, z are natural numbers.

Consider $1<x\leq y\leq z$. The other solutions are obtained by permuting the solution for this case. So it is sufficient to find solutions for which

$$1>\frac{1}{x}\geq \frac{1}{y}\geq \frac{1}{z}$$

Observe that $x\leq 3$. Indeed, if we assume x > 3 we get

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$$

If x=3 then obviously y=z=3. If x=2 then we have to solve

$$\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$$

Observe that $y \le 4$, otherwise we would have

$$\frac{1}{y}+\frac{1}{z}<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Thus for y=4 we obtain z=4, and for y=3 we obtain z=6.

To summarize, the equation has 10 solutions: six obtained by permutations of the numbers (2, 3, 6), three by permutations of the numbers (2, 4, 4), and (3, 3, 3).

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sat Mar 2 10:54:46 CST 2002