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In Memoriam
Mar 19, 2012
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?

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
But... but... you changed the numbers in your solution. :eek:

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:
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 \\

We can simplify this to:
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} \\

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.
Last edited:


In Memoriam
Mar 19, 2012
I have a good memory, but it's short(Nerd)