This is one of the puzzles included by Martin Gardner in a book containing various puzzles.-
Imagine that you have 3 boxes one containing 2 black marbles,one containing 2 white marbles,the 3rd containing one black and one white marble -labelled BB,WW, and BW respectively.Someone switched the labels so that every box is incorrectly labelled.You are allowed to take one marble at a time out of any box without looking inside and by this process of sampling you are to determine the contents of all 3 boxes.What is the smallest number of drawings needed to do this?
An equation of the type p^2-Nq^2=1 has been classified as Pell equation though it did not evolve from Mr Pell.
Given any value of N,one can find the values of p and q which satisfy this equation.
I have done extensive research on the subject ,finding the values of p and q for very many values of N-almost all upto 99 and some more values above 100.
There is a method for finding the solutions and this was used by me.
For values of N just above or below a perfect square the procedure does not require many steps but for values like 53,73 of N more steps were needed
As a variation for some values of N,you can get a solution for the alternative equation p^2-Nq^2=(-)1 after a few steps but will need more steps for p^2-Nq^2=1.
N=24-solution p=5 q=1,more solutions can be found later like p=49,q=10
N=53...p=66249,q=9100...This number N has solution p=182 q=25 for the alternative equation involving the final result as (-)1
At one time the Canadian and US dollars were discounted by 10 cents on each side of the border i.e a Canadian dollar was worth 90 US cents in the US and a US dollar was worth 90 Canadian cents in Canada. A man walks into a bar on the US side of the border,orders 10 US cents worth of beer ,pays with a US dollar and receives a Canadian dollar in change.He then walks aross the border into Canada,orders 10 Canadian cents of beer,pays with a Canadian dollar and receives a US dollar in change.He continues this throughout the day, and ends up with the original dollar in his pocket.Who has paid for the drinks he has consumed?
This is a rejoinder regarding the issue of 5 children weighing themselves two at a time to reduce the ..expenses.The weights revealed by the weighing machine in ascending order were 51.7 52 53.5 54 55 55.3 55.8 56.7 57 and 58.5.You were asked to determine the individual weights
If the individual weights were a,b,c,d and e the combinations 2 at a time possible are a+b,a+c,a+d a+e,b+c,b+d,b+e,c+d,c+e and d+e Adding all 10 values will result in the sum 4(a+b+c+d+e).So a+b+c+d+e=(51.7+52+53.5+54+55+55.3+55.8+56.7+57+58.5)/4=549.5/4=137.375.Deducting a+b=51.7 and d+e=58.5 we get c=27.175.We can then find that a=24.825,b=26.875,d=28.675 and e=29.825