Dimensions-Rectangular cuboid box-rejoinder

This is with reference to my earlier blog where the dimensions of a box are x y and z and these satisfy 3 equations and I wanted you to find the volume of the box.
Multiplying both the right and left hand sides of the 3 equations we get
xyz+x+y+z+1/x+1/y+1/z+1/xyz=4*7/3*1=28/3
But adding both sides of the equations we get x+1/x+y+1/y+z+1/z=4+7/3+1=22/3
Deducting result 2 from result1 we get xyz+1/xyz=2
This will be possible only when xyz=1 and so xyz the volume of the box is 1  

Dimensions-rectangular cuboid box

The dimensions of a rectangular cuboid box are x ,y and z. They satisfy the following equations-
x+(1/y)=4....y+(1/z)=1....z+(1/x)=7/3.
Find the volume of the box 

Puzzle-Syrup water mixture-Rejoinder

This refers to my previous blog wherein I had posted a question-From a container containing a syrup-water mixture in the ratio 5/3,how much should be removed and replaced by an equal quantity of water so that the syrup and water are contained equally
If  the syrup content is to be changed from 5/8 to 1/2 we have to remove 1/8 content of syrup which will need 1/5 of the total mixture--{5/8}-{(1/5)*(5/8)}=5/8-1/8=1/2.
The water content then will become 3/8-(1/5ths of 3/8)+1/5=3/8-3/40+1/5=    that is (15-3+8)/40=20/40=1/2 as required.
Other variations  ratio 5/7..1/7ths of the mixture to be replaced
                                    5/9..2/9 ths of the mixture to be replaced
                                    5/11..3/11 ths of the mixture to be replaced as follows
syrup 7/12-1/7 ths of 7/12=1/2  ..water   5/12-(5/12*1/7)+1/7=(35-5+12)/84=42/84=1/2
         9/14-2/9 ths of 9/14=1/2                 5/14-(5/14*2/9)+2/9=(45-10+28)/126=63/126=1/2
         11/16-3/11ths of 11/16=1/2            5/16-(5/16*3/11)+3/11=(55-15+48)/176=88/176=1/2

Puzzle-Syrup,water mixture

A vessel contains a liquid 3 parts of which are water  and 5 parts syrup.How much of the liquid must be drawn off and replaced with water so that the liquid may be half water and half syrup?
Alternately if the original liquid contains 5 parts water and 7 parts syrup?.

Sequence puzzle-rejoinder

This is a rejoinder to my blog posted on 11 june.-reproduced below-
A sequence of numbers a(1),a(2),a(3)......a(100) has the property that for every integer k between 1 and k inclusive the number a(k) is k less than the sum of the other 99 numbers.Find the 50th number-
There has been no response so far.
The solution can be arrived at as follows-
If S is the sum of the 100 numbers,a(k)=S-a(k)-k.
So 2a(k)=S-k
Substituting the values 1 to 100 for k  in this equation and adding all the results we will have
2S=100S-(the sum of the 100 numbers from 1 to 100) or
98S=5050 or S= 5050/98.
So a(50)=(5050/98)-a(50)-50
or 2a(50)=(5050/98)-50=150/98
or a(50)=75/98

Sequence puzzle

A sequence of numbers a1,a2,a3,....... a100 has the property that for every integer k between 1 and k inclusive,the number ak is k less than the sum of the other 99 numbers.Find the 50th number of the sequence.

Number in different forms

A person chooses a 3 digit number of the form N=ABC where A,B,C  are distinct digits.He then considered all possible numbers of the form ACB,BCA,BAC,CAB and CBA.He then added these 5 numbers  and got the result  3194.Determine the number N 

Page numbers of a book-Rejoinder

There has been no response so far.
However the solution can be arrived at as follows-
If there were 63 pages in the book,the sum of the page numbers from 1 to 63 will be 2016.
As the addition of an extra page number gave the total 2014,the total number of pages could not be 63 but 62.
For 62 pages the sum of the page numbers will be 1953 and so the page number added wrongly should be 2014-1953=61. 

Security passwords/numbers

Here is a puzzle regarding security code-
Before the entry gate of a secured post,the security guard will give a password to the person wishing to enter.The person has to give a reply.Only if the reply is correct he will be allowed to enter.
Person A comes to the gate.The guard mentions the number 12,A gives the answer 6. He is allowed to enter.
Person B comes to the gate-The guard mentions 6.B gives the answer 3.He is allowed to enter.
Person C comes to the gate after seeing the previous 2 persons entering.The guard mentions 8.C gives the reply 4.He is not allowed inside.
Can you make out why?
The code followed is this-When the guard gives a number,the person wishing to enter should find out the number of alphabets required for writing out the number and give it as the answer-6 letters for 12 and 3 letters for 6. You can see that for the number 8,it should be 5 instead of 4 due to which C was not allowed inside.

Page numbers of a book

A puzzle from Monday Edex issue (New Indian Express) of  9 june 2014-
The pages of a book are numbered 1 to n.When the page numbers were added by someone just for fun,one of the page numbers were added twice,resulting in an incorrect sum of 2014.Find the page number which was wrongly added twice..

Date of death-paradox

See the paradox below-
Both great writers Shakespeare and Cervantes died on the same date 23 April 1616.However they also died 10 days apart.How?
Here is the explanation-
Shakespeare died on 23 april 1616 according to Julian Calendar which was still used in England at the time whereas Cervantes died on 23 april 1616 according to the Gregarian calendar which was used in Madrid.The difference between the calendar accounts for the paradox.Spain had opted for Gregarian calendar as early as in 1582,while England waited till1752 to effect the change.

Who won what medals-Logic puzzle

Students of a school are divided into 3 teams-A,B and C
Using the clues given below,work out the number of gold,silver and bronze medals that A,B and C-
won-
1)A won 1 gold medal more than B,but 3 silver medals less than B
2)C has the highest number of bronze medals(18),but least number of gold medals(7)
3)Each team won at least 6 medals each type
4)B won  2 bronze medals more than gold medals
5)The 3 teams won 38 bronze medals in total
6)B won 27 medals in total
7)C won twice as many silver medals as the number of gold medals that B won

A number sequence to be filled-Rejoinder

I had asked readers to find the next number in the sequence 10,9,60,90,70,66
I had also suggested to treat this as an alpha-numeric sequence .
The numbers given are the number of alphabets which are required for expressing the highest numbers with  3,4,5,6,7,and 8 alphabets.
You are thus required to find the highest number which needs 9 alphabets for expression.
No one has given a solution.
Obviously the highest number has to be below 100.
Let us first consider numbers from 90 to 99.
97,98 and 99 require more than 9 alphabets.Only the number 96 requires 9 alphabets and so this is our candidate.  

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.