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

Solve for , 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 are all positive integers.

Then, if then which cannot be true for
positive finite integers, so . Now again, if , then are all & the sum of is at most . So .
Therefore =2, 3, or 4.

Case 1.

Let = 2.
This mean, then and are both , and the sum of is at most .
So .
Therefore, = 3, 4, 5, or 6.
Case 1.1 Let .
Then, , & .
Using reasoning similar to the above,
we have .
Therefore, = 7, 8, 9, 10, 11, or 12

Arranging the reciprocal sum and solving for ,
we have .
Plugging in the possibilities for , we find
the following solutions for :

Case 1.2 Let .
, & .
Reasoning as above, .
Therefore, = 5, 6, 7, or 8.
And .
Solutions are:

Case 1.3
Let
, and
Reasoning as above, .
Therefore, = 5 or 6.
and .
Solutions are:

Case 1.4
Let
and and
Reasoning as above, .
Therefore, =6.
and .
Thus the Solutions are:

Case 2: Let .
, and
Reasoning as above,

Therefore, =3 or 4.
Case 2.1 Let
and then , and
Reasoning as above,
Therefore, = 4, 5, or 6.
And, .
Solutions are:

Case 2.2 Let
and .
Reasoning as above,
Therefore, .
And
Solutions are:

Case 3
Let .
And , and
Reasoning as above,
Therefore, .
Case 3.1
Let then the Solutions are:

Summary of solutions:
.