Pigeon half, five-for-duck, quarter a-sparrow

We should feel sorry for our grand-children these days.

No nursery rhymes, no bedtime stories, no playground, no surprising of a crow trying to steal into the kitchen, no fishing and a cooling dip in the Royal Lake, and a lot more of such fun. Alas their moms are worse than drill sergeants. Saving money for the cramming sessions for their children next year or scouting for the best talent as cramming masters. Pure robotics!

Cramming for the exams, even for the KG!

All work and no play that's for you idiot. All play and no work is just for them, it's your karma. It's written for us.

My mother was something of a stern woman but when I grew up and made a living, I often wished I had a bit of her strength of character. And I didn't quite remember her telling me stories, though two of my big sisters did. The old mom-tortoise, son-of-papa-pelican-why-won't-you-go-to-sleep, little chick and the old tom-cat, golden rabbit and golden tiger went gathering thatch were the favorite stories of those times. Then there were also the riddles and puzzles and I happen to remember only this one:

"Pigeon half, five for duck, quarter a-sparrow,

            With 25 kyats, go get 25 birds in a row."

The problem is this. One duck costs 5 kyats, one pigeon costs 1/2 kyat, and one sparrow costs 1/4 kyat. Get 25 birds with 25 kyats. I vaguely remember getting to know it as a schoolboy and may be from friends, or my sisters. I am not sure if I heard of any solution or if I tried solving it myself. These days I started learning R and about two years ago, I solved it with some awkward R programming. It was a brute-force program, with a series of loops, just the kind of thing the R gurus looked down. The solution is: 3 ducks, 18 pigeons, and 4 sparrows.
Recently, I've been reading hurriedly about ancient Indian and Chinese mathematics and came across a class of equations known as indeterminate equations. The modern forms of such equations are Diophantine equations, Pell's equation, and the like. Absolutely I know just these few names and nothing more. If you are interested in these topics a good start may be the Wiki entry "Diophantine equations" and Wolfram MathWorld's entry with the same name.
Yesterday, while browsing for ancient Chinese mathematics on π related information, I found The Ambitious Horse: Ancient Chinese Mathematics Problems. There I found the 100 fowls problem. It reads:

"A rooster costs 5 copper coins call qian. A hen costs 3 qian, while 3 chicks can be purchased for 1 qian. You have 100 qian to buy exactly 100 fowls. How many roosters, hens, and chicks  there?"

What a coincidence, our duck-pigeon-sparrow puzzle is exactly the same kind. Then, is our 25 birds puzzle our own? Or was it an adaptation of this 100 fowls problem included in the Chinese classic Zhang Qujian suanjing of fifth century AD?
Or did we derive our puzzle from the Indian tradition of mastery of indeterminate equations from Aryabhata (499 AD)? This is not so far fetched if we recall that we have an old astrological tradition that has its roots in Indian astronomy. We now know that the first explicit description of the general solution in integers of the Diophantine equation ay - bx = c was found in Aryabhata's text called Aryabhatiya. His method was known as the kuttaka (pulverizing or breaking up into small pieces) method.  Aryabhata applied his method to solve simultaneous Diophantine equations for important applications in astronomy.
As for the 100 fowls problem, it is known that in problems such as these where the number of equations are less than the number of unknowns, there usually is more than one solution. Additionally we need a solution with integer values only.
Now the equations of the 100 fowls problem could be written as:

          5R + 3H + (1/3)C = 100,  and  R + H + C = 100,

where R is the number of roosters, H is the number of hens, and C is the number of chicks.

This is an indeterminate equation as there are three unknowns R, H, and C and just two equations. You could make them simpler by solving for C in the second equation,

C = 100 – R - H

and substituting for C in the first. Which will give you

           H = 25-(7/4)R.

But since you need solutions in whole numbers, you would create a new variable t = R/4, then

           H = 25 – 7t, and you can go on to find H using t = 0 or 1 or 2 or 3 or ...

From that you can find the possible pairs of values for H and t. R is then calculated as R = 4t. Then you get C from C = 100-R-H. Then you will get the four solutions, among which solution-2 is the Zhang Qujian's classic soluation:

                   R    H     C   Qian

Solution-1    0   25   75     100

Solution-2    4   18   78     100

Solution-3    8   11   81     100

Solution-4  12    4    84     100

If you don't allow leaving out the roosters, you get three solutions. Now you can try the same method with our 25 birds puzzle and there you will find it has only one solution. Here is my R program for the 100 fowls problem and it's a little more elegant than my brute force program mentioned earlier.