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.

Arithmetical progressions-3

I had posted 2 blogs earlier on the subject-the first for familiarising the topic-arithmetical progressions and second for finding subsequent terms/sums of terms etc.
Let me now show you an interesting observation-
Consider the 10 terms in the following progression-
1,5,9,13,17,21,25,29,33,37
Step 1-I will now multiply the first with the last,the second with the penultimate and so on and tabulate the results-
1*37=37..,5*33=165...9*29=261...13*25=325...17*21=357   viz 37,165,261,325,357
Step2- I start finding the differences between successive values above starting from the last 357
357-325=32,...,325-261=64....261-165=96...165-37=128  viz  32,64,96,128
Step3-Again I find the differences.I will get all values equal to 32.
Similar procedure has been adopted for the 8 terms in another progression  1,4,7,10,13,16,19,22
to get the following results
Step1-22,76,112,130
Step 2-18,36,54
Step 3- All values equal to 18.
You can get similar results for any arithmetical progression,starting with any number,involving any common difference,and any number of terms-only the number of terms should be even and not odd,as in that case there will be a middle term which will have no companion for multiplication
You can predict what the final result in the final step will be in any such case.
For instance consider the progression starting with 7,having common difference 5 and a total of 32 terms.The final result at the final step will be-You will get all values equal to 50.This is 2 times the square of the common difference 5 viz 2*5*5=50,like in the earlier examples shown viz 2*4*4=32 and 2*3*3=18

Divisibility

Find the smallest perfect square larger than 100000 which is divisible by 392?
2)What will be the smallest if it should be larger than 150000?
3)What will be the smallest larger than 100000,if it should be divisible by 1323?

Repeating Sequence-Rejoinder

This is with reference to my earlier blog with the same title.
I had shown a sequence with a few numbers--0,0,0,1,0,2,1,0,3,2,1,4,0,5,3,2,6,1,4......By deleting each number like 0,1,2,3,4 etc etc first occurring in it,the same sequence will be revealed.
I have worked further on it and have found further digits upto the 143rd place.I have shown below the sequence in successive rows -each row starting with 0.
0..........................................................................................contains 1 number
0..........................................................................................contains 1.number.
0,1.......................................................................................contains 2 numbers
0,2,1....................................................................................contains 3 numbers
0,3,2,1,4...............................................................................contains 5 numbers
0,5,3,2,6,1,7,4......................................................................contains 8 numbers
0,8,5,3,9,2,10,6,1,11,7,4,12.................................................contains 13 numbers
0,13,8,5,14,3,15,9,2,16,10,6,17,1,18,11,7,19,4,20,12.........contains 21 numbers
0,21,13,8,22,5,23,14,3,24,15,9,25,2,26,16,10,27,6,28,17,1,29,18
                11,30,7,31,19,4,32,20,12,33................................contains 34 numbers.
The total no of items in each group is 1,1,2,3,5,8,13,21,34..These numbers are part of a sequence called Fibonnaci number sequence,where starting from the 3rd item the successive numbers work out as the sum of the previous 2 numbers.
The next group which should contain 55 items(21+34) has not been shown by me.This group also starts with 0 (like the previous groups)and contains 33 as the last item.
In the full sequence,the digit 1 occurs at places   4,7,11,18,29,47,76,123  etc etc.Call this group A
The digit 2 occurs at places 6,10,16,26,42,68,110 etc.etc.Call this group B
In both these groups cach number commencing from the 3rd works out as the sum of the previous 2 numbers..Likewise you can have groups for digits 3,4,5,6 etc etc.
A small digression  relating to digits/numbers occuring at various places-one sample-
In the full sequence the number 13 occurs at place 35 for the first time.
                                               18            at place 48
                                               26            at place 69
                                         and 39           at place 103.
You can see that the differences regarding the number and the places are 5,8,13 and 13,21,34 which are numbers in the Fibonnaci sequence.The next numbers in this sequence are 55 and 89.
So you will find that the number 60 (which is 21 places above 39) will occur at place 158(which is 55 more than 103).Similarly 94 (which is 34 places above 60)will occur first at place 247(which is 89 more than 158). 

Prime numbers-in several forms

Take the number 1373
This has an interesting feature
Take each digit,or take two digits at a time in the same order,or 3 digits at a time in the same order or all 4 digits together they are all primes viz
1,3,7,3
13,37,73
137,373
1373 -all are primes
Can you find some more 4 digit numbers like this?