Mind-Boggling Puzzle: Solve How Many Parts Each Worker Got

[Wilmer]

Found this challenging:

A,B and C work at same speed.
When all 3 of them plant a field with D, the job gets done in 5 hours.
When all 3 of them plant the same field with E, the job gets done in 6 hours.
The field was divided between the 5 workers in proportion to their 5 work rates,
into a 2digit square number of parts, with E getting only 1 part.
A, B, C and D all got an integer number of parts.
How many parts did each get?

 

[Wilmer]

Well, that was a puzzle I found here:
Sunday Puzzle-19 - Welcome to the official website of Natalia Pogonina!

I changed the wording; answer same.
See my answer at bottom.

 

[I like Serena]

But... but... you changed the numbers in your solution.

Here's my solution.

Say $A,B,C,D,E$ are the number of parts each gets.
Let $a,b,c,d,e$ be their respective work rates in parts per hour.
And let $P$ be the square 2-digit number of parts.
Then it follows that:
\begin{array}{l}
A=B=C \\
a=b=c \\
A+B+C+D+E=P \\
E=1 \\
\frac Aa = \frac Bb = \frac Cc = \frac Dd = \frac Ee \\
5(a+b+c+d)=P \\
6(a+b+c+e)=P \\
\end{array}

We can simplify this to:
$$\left\{\begin{array}{l}
3A+D+1=P \\
\frac Aa = \frac Dd = \frac 1e \\
15a+5d=P \\
18a+6e=P \\
A,D \text{ whole numbers} \\
P \text{ square 2-digit number} \\
\end{array}\right.$$

By enumerating all square 2-digit numbers, we find $A=B=C=13,\ D=9,\ P=49$ as the only solution.
It will take them $\frac{234}{49} \approx 4 \text{ hours and }47\text{ minutes}$ to plant the field.

 

[Wilmer]

I have a good memory, but it's short(Nerd)