Recently I attended and magic function. The magician made a
very attractive proposal from the stage:
‘Can anyone in the audience give me Rs 5/ in 20 coins. One
condition. The coins must be of 50P, 20P and 5P denominations. No other coins
would do. To anyone who can give me this I am willing to give away Rs. 100/ one
hundred rupee for five!’
Everyone was silent. No one went forward. Some people began
to look for bits of papers and pencil in their pockets evidently to calculate
their chances. But no one went forward.
The magician renewed his offer once again. “What no takers.
No one wants to make easy money!’
There was silence in the auditorium.
‘Perhaps you think it is too much to give me Rs 5/- in
exchange of Rs 100/- Alright I’ll take only Rs 3/- Of course, in the same
denominations as I mentioned already. Twenty coins. How about that now?
No one stirred.
‘Alright, alright! The magician went on. ‘Even three rupees
you think is too much to exchange for Rs 100/ - I will come down even more.
Only two rupees – just two rupees’ he showed his two fingers ‘for rupees one
hundred’. You can’t let go of such an opportunity, really. Ladies and
gentlemen. Just two rupees – in the denomination I mentioned already – twenty
coins – for rupees one hundred!’
Nothing happened. He renewed his offer several times and
finally he gave up.
Why do you think no one came forward to take advantage of
the magician’s most attractive proposal?
Answer: All the
problems proposed by the magician are insoluble. The magician could easily make
such an offer, very well knowing that he never would have to part with his
hundred rupees. Let us now analyze the problem algebraically, to know exactly
where the magician had his safety valve:
To pay 5 rupees: Let us assume that it is necessary for us
to have ‘a’ number of 50P coins. ‘b’ number of 20P coins and ‘c’ number of 5P
coins. Then we will have the equation:
50a+ 20 b +5c = 500P = 5 rupees.
Simplifying this we get: 10a +4b +c = 100
However, according to the problem, the total number of coins
is 20, and therefore we have the other equation: a+ b+ c =20.
When we subtract this equation from the first we get: 9a +
3b = 80
Dividing this by 3 we obtain: 3a + b =26 ²⁄₃. But 3a i.e.
the number of 50P coins multiplied by 3 is, of course, an integeger like b, the
number of 20 p coins . And the sum of these two numbers cannot be a fractional
number. Therefore the problem is insoluble. In the same way the ‘reduced’
payment problem are also similarly insoluble. In the case of Rs. 3/ we get the
following equation: 3a x b = 13 ½
And in the case of Rs. 2/ we get the equation: 3a + b = 6
²⁄₃
Both as we can see are fractional numbers. Therefore the
magician risked nothing in making such a generous had he asked for Rs. 4/
instead of Rs. 5/ Rs. 3/ or Rs. 2/. Then we could have found seven different
solutions to the problem.
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