Filling up of a sequence-2

This is with reference to my earlier blog-same title.
In spite of my hint that the solution is in the question itself nobody has responded.
The question was-'Anybody willing to work out the next two numbers in the sequence below?'
You have to look at the number of alphabets in each word in the sentence above forming the question
The numbers of alphabets in successive words are 7,7,2,4,3,3,4,3,7,2,3.The last two words viz 'sequence' and 'below' contain 8 and 5 alphabets respectively and thus these two numbers 8 and 5 are the numbers filling up the sequence .

One set of faulty coins-2

This is a rejoinder to my earlier blog of same title.
A correct reply has already been given by Shekar which may be seen under the comment portion.
The reply holds good if the weighing machine is capable to weigh correctly  upto the weight of around 55 coins.Suppose it is capable to weigh correctly upto the weight of say around 45,46 etc of coins.In that case you must take no coins from one set,which you can mark.If all coins weigh correctly the total weight should be that of 45 coins,(45 being the sum of digits 1 to 9) and so the actual weight differring from weight of 45 coins will show you which set has the faulty coins.  

How to assign tasks-rejoinder

This is with reference to my earlier blog regarding the assignment of 6 tasks to 6 persons
Task 2 can be assigned only to either person 3 or 4..
As task 1 cannot be assigned to either person 1 or 2,it could be assigned to either persons 3,5,6 or 4,5,6 depending on 4 or 3 being assigned task 2-So Tasks 1 and 2 can be assigned only in 6 ways viz 4/3,4/5,4/6,3/4,3/5,3/6.
The remaining 4 tasks can be assigned in 4! or 24 ways
Thus the total number of ways in assigning the 6 tasks is 6*24=144 ways.  

Nomination for liebster blog awards

Thank you,Aparna for nominating me for the liebster award.
The following are  the answers to your questions
1)I would like to see all teachers respected by their students
2)I was in need of help for having  a darshan in a temple and someone not known to me helped me by leading me with his hands.
3)My keen interest in mathematics
4)Variety is not confusing if looked at properly
5)We must avoid attending to some other job like speaking on phone etc
6)Thanking
7)To age without suffering any illness
8)Every person must readily help others needing it
9)Eat well,work well and forget all problems
10)Globalised  .
As per the procedures generally  followed I have nominated the following 5 persons for the Liebster Blog awards.-
1)rajirules.blogspot. in
2)mind your decisions.com/blog
3)bhargavbalakrishnan.blogspot.in
4)Insane by thoughts and human by living
5)http://anushankar.blogspot.in
I wish to raise the following questions-
1)What will you do if you are alone in a lonely island
2)What happens when an aircraft loses fuel
3)Can you find anything special about the word 'undergrounder'
4)Will you count zero as an odd or even number
5)Do you like astrology
6)Which sea is colder-Arctic or Antarctic
7)Can a day have more than 24 hrs
8)Out of geography,physics,mathematics or history which do you like most
9)Is it possible to have a meteorite crash without serious damage to persons,buildings etc
10)Out of the following which would you not like to miss-a music concert,a cricket match ,a tv serial or the uptodate news from media
I introduce myself as a honours graduate in physics ,now retired from several jobs including a stint in Libya,having interest in music,mathematics and acquaintance with words in english language.
I will be intimating the persons whom I have nominated to follow this up further by nominating  other bloggers known to them and communicate their responses to my questions-the first coming to their mind.

Movement of digits

 Consider the number shown below-
523814769
This number consists of each of the nine digits from 1 to 9 without any repetition and it is a perfect square..
You are required to convert it to 123456789 by movement of digits from one place to another by interchange.
For example by interchanging 5 and 6 from 523814769 you get 623814759.This is considered as one movement.
You are allowed 3 movements for conversion of 523814769 into 123456789.
Proceed.

Alphabet/number link

The following is an addition sum where each alphabet represents a distinct digit from among the 10 digits from 0 to 9-
    S  E  N  D
    M O  R  E
 -----------
M O N  E  Y
Find the actual addition sum as per those digits.

Filling up a sequence-two numbers to be filled up

Anybody willing to work out the next 2 numbers in the sequence below?-
7,7,2,4,3,3,4,3,7,2,3,?,?-
The answer is in the question itself.

Conversation with priest-Rejoinder

The organist could find the alternatives available for the values of ages where the product works out as 2450.As the sum of the ages was to be twice his own age and as he was unable to give the answer in the first instance,we can conclude his age as 32 and there were  2 alternatives available as (49,10,5) and (50,7,7) After the priest mentioned that he was the oldest,the organist could give answer.The logic works out as follows-If 'p' is the age of the priest and if p>50,he could not give the answer.If p<50,he could not be the oldest.So the organist concludes that p=50 and accordingly the ages of the visitors were 49, 10 and 5. 

One set of faulty coins

There are 10 sets of coins.You know how much the coins should weigh.You also know that all the coins in one set of 10 are exactly a hundredth of an ounce off,making that entire set of 10 coins a tenth of an ounce off.You are also told that all the other coins weigh the correct amount.You are allowed to use an extremely accurate digital weighing machine only once.How do you determine which set of 10 coins is faulty?

How to assign tasks

There was an organisation which had at a time the work of assigning 6 tasks to 6 persons on their roll.Each had to be assigned one task.Task 1 cannot be assigned to either person 1 or 2.Task 2 must be assigned to either person 3 or 4.In how many ways can the 6 tasks be assigned to those 6 persons?

Conversation between a priest and an organist in a church

The priest told the organist'Today there were 3 persons in the church besides us.The product of their ages was 2450 and the sum equal to twice your age.How old were they?' The organist who was clever,thought for a while and said'No,I cant figure that out'.The priest replied'If I tell you that I was the oldest person in the church today,can you solve the problem?'.The organist was able to do that.
Now comes a question for you-How old was the priest?  

Solutions for x^3+y^3+z^3=2

If we have the equation x^3+y^3+z^3=2,will it be possible to have integer solutions?
A first look will make you think that there  are none,but there are infinite number of solutions
You put x=1+p and y=1-p.You can then work out z as well as the value of p.
x^3+y^3=(1+3p+3p^2+p^3)+(1-3p+3p^2-p^3)=2+6p^2.
So we must have z^3=(-)6 p^2.
If p=(-)6a^3 ,we can get z^3=(-)6*(36a^6)=(-)216 a^6 and so z=(-)6 a^2
Thus the solutions are x=1-6a^3,y=1+6a^3 and z=(-)6 a^2 for any value 'a' can take
Example if a=2,we have x=1-48=(-)47,y=1+48=49 and z=(-)24
x^3+y^3+z^3=(49^3)-(47^3)-(24^3)=117649-103823-13824=2.

One more cyclic number-Rejoinder 2

The link mentioned in the just preceding blog occurs as follows-

The numbers used for multiplication in the reverse order are 1,....621517,...269324,...441277,....251300,..25130,...2513,...1864802,..1429514,..557296,..1713108.

1000000,100000,10000,1000,100,10..

The number 2513 can be worked from the previous 6 numbers by successively multiplying them with 6,2,1,5,1,7 forming part of 621517 and adding the results.Similarly the number 1864802 can be worked from the previous 6 numbers etcetc-see below-

(1*6)+(621517*2)+(269324*1)+(441277*5)+(251300*1)+(25130*7)=

6+1243034+269324+2206385+251300+175910=4145959.From this result the maximum possible multiple of 2071723 viz 4143446 is deducted producing 2513.

Similarly (621517*6)+(269324*2)+(441277*1)+(251300*5)+(25130*1)+(2513*7)=6008248 which after subtracting 4143446 gives us 1864802.etc etc.

You can similarly get all the subsequent numbers.

One more cyclic number-Rejoinder

This is with reference to my post of 22 june 2012,where I had introduced a cyclic number of 17 digits with 6 zeros viz 00000048269001213-Call this 'N'.
I had mentioned that the 6 zeros occur due to the action of finding the reciprocal of  2071723-You can call this as'g'.
I had also mentioned that by multiplying the number 'N, with  621517, you will get a new number as follows-
30000004826900121-viz the 3 at the end of the original number 'N' moving to the left with all other digits remaining as before in the same order
I had not mentioned how this number for multiplication could be found.It is found as follows-
You have to multiply g=2071723 by a digit 'd'which gets a result of the form (10k-1)  and you will find that k=621517.
We have d=3 and the multiplication gives us 6215169 and so k=621517.
The subsequent numbers which on multiplication with 'N" produce results like 13000000482690012..
21300000048269001 etc etc-showing the last digits moving to the left successively are all the remainders at each stage while taking the reciprocal of g=2071723.
The procees of finding the reciprical gives these numbers as follows-
The first stage for getting a quotient while dividing 1 by g occurs after adding 7 zeroes to 1,making it
10000000-The division by 'g' gives a quotient 4 and remainder 1713108.
At the step dividing 17131080 by 'g' gives the quotient 8 and the remainder 557296.
Proceeding further we get quotients 2,6.9 etc forming part of 4826900121 .
All the remainders are 1713108,557296,1429514,1864802,2513,25130,251300,441277,269324 and finally 621517 and 1-The cycle repeats from here again.
These remainders when used in the reverse order for multiplication of 'N' will produce results showing movement of the last digits successively to the left as shown below-
621517 will produce 30000004826900121
269324 will produce 13000000482690012
441277 will produce 21300000048269001 etc .etc.
All these numbers used for multiplication also have a link in another way which I will show in my next post.  

Ramanujan number-2

In my previous blog of 18th feb,I had mentioned about the number 1729-called Ramanujan number-the only number which could be expressed as a sum of 2 cubes in 2 different ways-viz 1000+729=1728+1=1729.
If we consider differences between 2 cubes (this means the same thing as considerig the cube of one negative number),we can write the equations differently....12^3+{(-)10}^3=1729-1000=728 also 9^3+{(-1)}^3=729-1=728.
I can now show a case where a number can be expressed as a difference of 2 cubes in 3 different ways
34^3-33^3=39304-35937=3367
15^3-2^3 = 3375-8 =3367
16^3-9^3 = 4096-729=3367
There could be many more such cases.-also numbers which could be expressed in more than 3 different ways.

Sums of 2 square numbers-rejoinder

With reference to my 2 prvious blogs on the subject there are only 4 cases relating to numbers between 2001 and 3000.The following are those cases.

484+1784=2248,400+1849=2249,225+2025=2250
1296+1296=2592,2304+289=2593,1225+1369=2594
784+1600=2384,81+2304=2385,361+2025=2386
196+2116=2312,9+2304=2313,289+2025=2314.

I am yet to work about cases of numbers above 3000.

Sum of 2 square numbers-2

With ref to my blog of 9 march,no one has come with a response.The following are the 8 cases of numbers between 1001 and 2000
196+900=1096,256+841=1097,9+1089=1098
576+576=1152,64+1089=1153,529+625=1154
16+1296=1312,289+1024=1313,225+1089=1314
64+1600=1664,576+1089=1665,441+1225=1666
144+1600=1744,784+961=1745,225+1521=1746
900+900=1800,576+1225=1801,121+1681=1802
576+1296=1872,784+1089=1873,25+1849=1874
196+1764=1960,361+1600=1961,441+1521=1962.
I will show the cases of numbers between 2001 and 3000 in my next blog

Sums of 2 square numbers

There are not many cases of 3 consecutive numbers where all three of them can be expressed as a sum of 2 squares
There are only 7 cases with numbers below 1000,which satisfy this requirement.These are shown below with the squares involved in each case shown within brackets-
72(36+36)..73(9+64)...74(25+49)
288(144+144)..289(64+225)..290(289+1)..This  of course can be rejected as one of the numbers 289 is itself a square.
520(196+324)...521(121+400)...522(81+441)
584(100+484)...585(144+441)..586(225+361)
800(400+400)...801(225+576)...802(361+441)
232(36+196)...233(64+169)...234(9+225)
808(324+484)...809(25+784)...810(81+729).
Taking into account cases similar to 288,289,290 which have to be rejected,there are 8 cases of numbers between 1000 and 2000 and  5 cases relating to numbers between 2000 and 3000.Can you find them?