Few math problems, specially, problems on Numbers are very interesting. In this “Note”, I’ve added a classical problem, as follow:

Solve $\dfrac{1}{w} + \dfrac{1}{x}+ \dfrac{1}{y} + \dfrac{1}{z}=1$ for $w \le x \le y \le z$ , all being positive integers.

– This problem is quit easy to solve but interesting to understand steps, how it is solved. One with regular math knowledge would know that there are fourteen (14) solutions for the problem. Some where this problem is also called Egyptian Fractions Problem.

Solving Steps

Since {w, x, y, z} are all positive integers.

Then, if w = 1 then 1/x=1/y=1/z = 0 which cannot be true for
positive finite integers, so w > 1 . Now again, if $w \ge 5$ , then x, y, z are all $\ge 5$ & the sum of $\dfrac{1}{w} + \dfrac{1}{x}+ \dfrac{1}{y} + \dfrac{1}{z}$ is at most 4/5 < 1 . So w < 5 .
Therefore =2, 3, or 4.

Case 1.

Let = 2.
This mean, $2 \le x \le 6$ then and are both $\ge 7$ , and the sum of $\dfrac{1}{x}+ \dfrac{1}{y} + \dfrac{1}{z}$ is at most 3/7 < 1/2 .
So $x \le 6$ .
Therefore, = 3, 4, 5, or 6.
Case 1.1 Let x = 3 .
Then, $3 \le y \le z$ , & 1/y + 1/z = 1/6 .
Using reasoning similar to the above,
we have $6 < y \le 12$ .
Therefore, = 7, 8, 9, 10, 11, or 12

Arranging the reciprocal sum and solving for ,
we have $z =\dfrac {6y}{(y - 6)}$ .
Plugging in the possibilities for , we find
the following solutions for x=3 :
$\text {y=7, z=42} \\ \text {y=8, z=24} \\ \text {y=9, z=18} \\ \text {y=10, z=15} \\ \text {y=12, z=12}$
Case 1.2 Let x=4 .
$4 \le y \le z$ , & 1/y + 1/z = 1/4 .
Reasoning as above, $4 < y \le 8$ .
Therefore, = 5, 6, 7, or 8.
And $z= \dfrac{4y}{y-4}$ .
Solutions are:
$\text {y=5, z=20} \\ \text {y= 6, z=12} \\ \text {no solution for y=7} \\ \text {y=8, z=8}$
Case 1.3
Let x=5
$5 \le y \le z$ , and 1/y + 1/z=3/10
Reasoning as above, $5 \le y < 7$ .
Therefore, = 5 or 6.
and $z = \dfrac{10y}{3y-10}$ .
Solutions are:
$\text {y=5, z=10} \\ \text {no solution for y=6}$
Case 1.4
Let x = 6
and $6 \le y \le z$ and 1/y + 1/z = 1/3
Reasoning as above, $6 \le y \le 6$ .
Therefore, =6.
and $z = \dfrac{3y}{y - 3}$ .
Thus the Solutions are:
{y = 6, z = 6}
Case 2: Let w = 3 .
$3 \le x \le y \le z$ , and $\dfrac{1}{x}+ \dfrac{1}{y} + \dfrac{1}{z}= \dfrac{2}{3}$
Reasoning as above,
$3 \le x < 5$
Therefore, =3 or 4.
Case 2.1 Let x = 3
and then $3 \le y \le z$ , and 1/y + 1/z = 1/3
Reasoning as above, $4 \le y \le 6$
Therefore, = 4, 5, or 6.
And, $z = \dfrac{3y}{y-3}$ .
Solutions are:
$\text {y=4, z=12} \\ \text {no solution for y=5} \\ \text {y=6, z=6}$
Case 2.2 Let x=4
$4 \le y \le z$ and 1/y + 1/z = 5/12 .
Reasoning as above, $4 \le y < 5$
Therefore, y = 4 .
And $z= \dfrac{12y}{5y-12}$
Solutions are:
{y = 4 , z = 6}

Case 3
Let w = 4 .
And $4 \le x \le y \le z$ , and $\dfrac{1}{x}+ \dfrac{1}{y} + \dfrac{1}{z}= \dfrac{3}{4}$
Reasoning as above, $4 \le x < 5$
Therefore, x= 4 .
Case 3.1
Let x=4 then the Solutions are:
{y = 4 , z = 4}

Summary of solutions:
$\text{w= 2 , x= 3 , y=7, z=42} \\ \text{w=2 , x=3 , y=8 , z=24} \\ \text{w=2 , x=3 , y= 9 , z=18} \\ \text {w=2 , x= 3 , y=10 , z= 15} \\ \text{w=2 , x=3 , y=12 , z=12} \\ \text{w=2 , x=4 , y= 5, z=20} \\ \text{w=2 , x= 4 , y= 6 , z = 12} \\ \text{w=2 , x=4, y=8 , z= 8} \\ \text{w = 2 , x=5 , y=5 , z=10} \\ \text{w=2 , x= 6 , y=6 , z=6} \\ \text{w=3 , x=3 , y= 4, z= 12} \\ \text{w=3 , x=3 , y=6 , z=6} \\ \text{w=3 , x=4 , y= 4 , z= 6} \\ \text{w = 4 , x= 4 , y=4 , z= 4}$ .

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Solving Steps

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