Trilemmas

The term trilemma is similar to the term dilemma.
It refers to 3 choices/statements available out of which only two can be seen as true or amenable to acceptance
There are many examples-I have shown 2 of them below-
1)Manufacturing processes-Choices available-Fast...Good....Cheap
2)Three statements regarding religious beliefs
a)If god is unable to prevent evil,then he is not very powerful
b)If god is not willing to prevent evil then he is not at all good
c)If god is both willing and able to prevent evil then why does it(evil) exist?

Decimal fractions puzzle-Rejoinder

Though it is a very easy puzzle, no one has given a reply so far.
Let me explain the method.
For the 3 fractions x,y and z ,I had mentioned that x(1),y(1) and z(1) are the integer parts and x(2),y(2) and z(2) the decimal parts.
So you can see that x=x(1)+x(2).....y=y(1)+y(2)     and z=z(1)+z(2).
I had given 3 equations x+y(1)+z(2)=4.2
.                                  y+z(1)+x(2)=3.6 and
                                   z+(x(1)+y(2)=2.0
Adding all 3 together we get x+y+z+x(1)+y(1)+z(1)+x(2)+z(2)=9.8 or because x=x(1)+x(2) ..etc
we have 2(x+y+z)=9.8 or x+y+z=4.9
Deducting the first equation from this we will have x+y+z-x-y(1)-z(2)=4.9-4.2=0.7 which means
y(2)+z(1)=0.7...This further means that z(1)=0 and y(2)=0.7
Similarly deducting the 2nd and 3rd equations we will get x(1)=1 and z(2)=0.3 and y(1)=2 and x(2)=0.9
Thus the 3 fractions are 1.9,....2.7 ... 0.3.

A number sequence to be filled

You are required to find the next number in the sequence 10,9,60,90,70,66
You will find it difficult to find the number if you tackle it as a mathematical issue
The actual way is to tackle it as an alphanumeric issue.
10 is the highest number which can be written in 3 alphabets
9 similarly the highest which can be written in 4 alphabets.
Likewise 60,90,70,66 -those which are highest as could be written in 5,6,7 and 8 alphabets.
Now go ahead and find the highest number which could be written in 9 alphabets.

Decimal Fractions Puzzle

Let x,y and z be 3 decimal fractions.The integral portion in each is denoted by x(1),y(1) and z(1) and the fractional portion by x(2),y(2) and z(2) respectively.
If x+y(1)+z(2)=4.2,y+z(1)+x(2)=3.6 and z+x(1)+y(2)=2,Find the values of  all the 3 fractions..

Divisibility-Rejoinder

My previous blog on this topic dealt with squares and their divisibilities-
1)About the square numbers divisible by 392-
First find the nearest square divisible by 392-It is 2*392=784(square of 28)
You must now multiply 784 by any other square number to find all squares divisible by 392-
like 4,9,16,25,36,49 64,81,100,121,144,169 etc to find the nearest square above or below any limit.
Multiplication of 784 by 144 gives you the nearest square above 100000,while multiplication by 121 gives you the square nearest below 100000
2)Multiplication of 784 by 225 gives the nearest square above 150000
3)Regarding divisibility  by 1323,we find its factors as 3,3,,3,7, and 7.So you must multiply by 3 to make it a square viz 3969-Square of 63.You can then decide which square you have to  multiply with 3969,to find the squares above or below any limit. 

Puzzle given to me by Vidyu

A puzzle was given to me by Vidyu as follows-
There is a four digit number 'aabb' formed by the 2 digits'a' and 'b' which is a perfect square.Find the values of 'a' and 'b'.
There are no such cases in our normal 'decennial' system.i.e numbers to base 10.
It could be in a system with another base.
If the base is 'x',the number can be expressed as ax^3+ax^2+bx+b which can be simplified as
(ax^2+b)(x+1).
If this has to be a perfect square both (x+1) and (ax^2+b) should be squares.
Thus x can have a value 3 or 8.(See below where x=3)
If x=8 we must have (64a+b) must be a square.The only possible values for a and b are a=5 and b=4.as 64*5+4=324 which is a square.
Thus the solution is 5544 in base 8.
5544 in base 8 is equal to 2916 in base 10 which is the square of 54(base 10)
54 in base 10 is equal to 6*8+6 i.e 66 in base 8.
Thus purely considering both numbers in base 8,...5544 is the square of 66
You can check by multiplying 66 by 66 following the rules for all calculations viz multiplication,addition etc for base 8.
6*6=36=4*8+4=44 in base 8.We have to carry forward 4.
6*6+4(carried forward)=36+4=40=50 in base 8.
Thus 6*66=504 and so 66*66=5544.
There is no use in considering the value x=3 as in that case the number should be 0011,which is not a 4 digit number.
As a matter of variation, I cosidered other bases.
In base 9 ,the square of 66 will be 4840  and in base 7 it will be 6501,You can check. 

Tossing of a coin-Probabilities

Suppose you toss a coin-You may get a head(H) or tail (T)---2 variations
Suppose you toss it a second time-You may get
 HH,HT,TH,TT-4 variations
Suppose you toss once more-
You may get HHH,HHT,HTH,HTT,THH,THT,TTH,TTT-8 variations
Suppose you try to find out the number of cases where 2 heads do not occur in consecutive tosses
You will find the following position-
2 tosses-HT,TH,TT---3 cases
3 tosses-HTH,HTT,THT,TTH,TTT-5 cases
If you procced further you will find the following-
4 tosses-8 cases
5 tosses-13 cases
The number of cases of this occuring are respectively 3,5,8,13....These are numbers in the famous Fibonnaci sequence where each number from the 3rd is the sum of previous 2 numbers
The further number of cases are thus 21,34,55,89,144...etc for 6,7,8,9,10...number of tosses.
Of course the numbers of cases where 2 tails occur in consecutive tosses are also the same.