In ancient India, it appears that there were public competitions for the solving of difficult problems. Although this may seem surprising, we do know that the Hindus made a significant contribution to the mathematics of their time. Because of their lack of algebraic notation, the Hindus described their problems in words. They used great imagination in ‘decorating’ their problems, but were less concerned with realism! Here is a Hindu problem.

A group of bees equal in number to the square root of half the whole swarm alighted on a jasmine bush, leaving behind 8/9 of the swarm. And only one little bee circled about a lotus, for it was attracted by the buzzing of a sister bee that was so careless as to fall into the trap of the fragrant flower. How many bees were there in the swarm?

Note: There is an ambiguity in this old problem. You will need to assume that the two stray bees do not belong to the remaining 8/9 of the swarm.

Hint 1

Make this into an algebraic problem.  For example you might let S be the number of bees in the swarm..

Hint 2

Remember that we are looking for solutions in integers.

Solution

Ambiguities should never occur in the statement of problems, but human nature being what it is ... !

As suggested in the hint, we let S denote the number of bees in the swarm. Now the number of bees in the group which alighted on the jasmine bush is √(S /2). This leaves  8/9  of the swarm plus the two stray bees. The total of all these bees is, of course, just the number of bees in the swarm itself.

Hence we obtain the algebraic equation:

√(S /2) + 8S/9 + 2 = S,
or
9√(S /2) = S – 18. (*)

How do we solve this equation?  Since we want an integral solution, S must be even (to make the left hand side integral) and S > 18 (to make the right hand side positive). To make the square root integral we must have S /2 = (1, 4, 9,) 16, 25, 36, 48, ... .  That is, S = 32, 54, 72, 98, ... .  Substituting these values in turn in (*), we obtain for S = 32: 36 > 14; for S = 50: 45 > 32; for S = 72: 54 = 54; for S = 98: 63 < 80; ... . Assuming (or proving, for example using calculus) that the right hand side continues to increase faster than the left hand side, this shows that S = 72 is the unique solution.

An alternative is to square both sides of equation (*) and simplify to obtain the quadratic equation

     2S²  –  153S  +  648  =  0.

This can now be solved using ‘the formula’, or we can observe that 648 = 2 x 2 x 2 x 3 x 3 x 3 x 3. Since the solution S is an integer, S must divide 648 exactly; hence S must occur as the product of some or all of these factors. Trial and error now gives  S = 72  as before.

Extensions

1. Given that N = 144 is the unique positive integral solution of

       √N + N / 2 + 60 = N,

make up a story which leads to this equation and solution.

2. Invent some other problems of this type.

Hint 1

Hint 2

Solution

Extensions