Welcome to PUZZLE MANIA(NOV 3rd - NOV 8th)

OVERALL WINNERS OF PUZZLE MANIA

RESULTS OF PUZZLE MANIA - puzzle 6 *NEW

PUZZLE 6 - SOLUTION

PUZZLE 5 - SOLUTION
PUZZLE 4 - Solution
PUZZLE THREE ------ SOLUTION
PUZZLE TWO ------ SOLUTION ~ ~ ~ ~ ~ PUZZLE ONE ------ SOLUTION

RULES:

The rules of the event are simple. This is not a team event and therfore all the participants are advised to send their answers through their unique Email ID. Participants including those of NIT Durgapur please do not forget to write your name, year , branch and college name when you send the answers to mathsntech.nitdgp@gmail.com ,When you submit the answer through mail write 'PUZZLE 1 ANSWER' in the subject.One candidate is allowed to use only one e-mail ID throughout the competition. If multiple ID's are found, then the candidate will be disqualified.

OPENING & CLOSING TIME:

A puzzle will be put up daily all through the week at 10 P.M starting November 3rd '08 to November 8th '08. Solve the puzzle and then submit the answers to mathsntech.nitdgp@gmail.com before 8 P.M on the next day. This means entries to PUZZLE 1 will start at 10 P.M on MONDAY and end at 8 P.M on TUESDAY.

DECLARATION OF WINNERS :

Mere submission of answers doesnt ensure your win. Only the fastest fingers win. Take care that in your haste to submit the answer you dont miss out writing the explanation. Incomplete explanation will be treated as wrong answer. If you happen to be in the first 10 correct entries your name will be put up on the winners list along with your college name. There are gift vouchers for Daily Puzzle Winners and Cash prize for the Overall winner which will be declared on sunday !!!

Discussion Forum :

You are free to discuss in the comments forum in the blog page. This space will be uploaded every 30 minutes. However direct posting of the answer will be rejected. Anonymous comments will also be rejected.

The decision of Maths N Tech Club Team is final and no queries will be entertained regarding the declaration of the winners.

Happy solving. Wish you all the best !!!

WINNERS OF PUZZLE MANIA

WINNERS OF PUZZLE MANIA:

Winners of Puzzle Mania's first edition were chosen on the following basis : Maximum Right Answers (in case of a tie puzzle 6 is divided as 2 puzzles) - Fastest timing - Proper Explanation.

The following students are on this basis come in the top 10 List of total PUZZLE MANIA. Many Congratulations to them. There is cash prize for the OVERALL WINNER and OVERALL RUNNER. There are also gift vouchers for the winners of individual puzzles. The overall winner will be getting a Certificate of Merit from Maths 'N' Tech Club.

Thank you for participating in the first edition of puzzle mania. We welcome your participation in the second edition too planned in March 2009. Any feedback is welcome at mathsntech.nitdgp@gmail.com



Congratulations Sayanta Sundar Bhar - Winner, PUZZLE MANIA
Congratulations Dipesh Kumar Kalyani - Runner, PUZZLE MANIA


The following students will recieve a mail from Maths N Tech Club in the next 2 days. You will be informed regarding the collection of the prizes :

PUZZLE 6 SOLUTION.

PUZZLE 1:
From the given condition of MM/DD/YYYY, we can conclude that the next palindrome would occur only in the 13th Century. I recieved many answers as 12/31/1321. But that is not the answer. We have a palindromic date before it which is in 1390.

People who have thought about this year made another mistake, they put down the date as 09/31/1390. But they missed out on the fact that the 9th month ie september has only 30 days. So this date is not valid and the correct answer turns out to be 08/31/1380. which is the CORRECT ANSWER.


Puzzle 2:
Singh -> L R L R L R L R L
kING -> L R L R L

THUS ONLY LL PAIR GETS REPEATED, I.E. A CYCLICITY HAS OCCURED. HENCE BOTH SETTING THEIR RIGHT FOOT IS IMPOSSIBLE. INFINITE STEPS IS ALSO CONSIDERED AS THE RIGHT ANSWER.

Only those entries were considered right which had both of them correct. However in deciding the overall winner solving atleast one puzzle right will give the advantage.

PUZZLE 5 SOLUTION

The solution to puzzle 5 is as follows :

PUZZLE 6 - DOUBLE TROUBLE

Today is the last day of puzzle mania. Today's puzzle consists of 2 questions. You need to answer both of them to contend for the winner's slot.




OVERALL WINNER OF PUZZLE MANIA WILL BE ANNOUNCED TOMORROW 9 PM.

PUZZLE -5 THE STRANGE SERIES

Your Answer should be sent in the following format :
To : mathsntech.nitdgp@gmail.com
Subject : Puzzle -4 Answer
Answer :

Solution : (Explaining how you reached at the solution)
Name , Year , Branch , College !!!

PUZZLE 4 SOLUTION

First consider the shortest distance that is between ECE and CIVIL = 4 Km. Therfore you can have only 2 roads between and the only possible way of doing this is (1+3) or (3+1) kms. And that means it has to pass through IT. Now it is given that the CIVIL-CHEMICAL road is 12 kms long. If CIVIL -IT is 1, then CHEMICAL - IT can be maximum 10, So minimum distance comes out to 11 Km which is contradictory. Therfore it has to be 3 kms. Now we know CIVIL-IT is 3 kms and IT-ECE is 1 km. Here onwards we can continue the logic and reach the following conclusion.

The correct match is :
ECE - IT - 1 km
Mech - Elec - 2 km
CIVIL- IT - 3 km
ECE - CSE - 4 km
ELEC- CHEM -5 km
ELEC - IT - 6 km
ELEC- CIVIL - 7 km
MECH - ECE - 8 km
ELEC-CSE - 9 km
IT - CHEM - 10 km

PUZZLE - 4 - The Distant Roads

NIT Durgapur is to be upgraded into IIT Durgapur. For this the ministry of HRD called for development of the college infrastructure including construction of new buildings for the following branches of the college - Mechanical, Electrical, Civil, Chemical, CSE, ECE, and IT. These new buildings were then connected among themselves with 10 roads. The figure below shows the sketch of the pattern in which they were constructed and connected with roads.

The speciality of the roads is that each of the 10 different paths have a diffeent length between 1 Km - 10 Km. Some information is given in the table regarding the roads. The numbers in the table indicate the shortest distance between the two departments. For example, The shortest distance between chemical and CSE is 14 Kms. Also the CIVIL- IT and the ELECTRICAL - CHEMICAL roads are non-intersecting.

Find out the lengths of each of these 10 roads.

Note: Answer will be considered complete if all the roads are correctly matched with the lengths.

Your Answer should be sent in the following format :
To : mathsntech.nitdgp@gmail.com
Subject : Puzzle -4 Answer
Answer :

Solution : (Explaining how you reached at the solution)
Name , Year , Branch , College !!!

Happy solving !!!

PUZZLE -3 SOLUTION

I believe many of you have made mistake in estimating the total volume affected by the cubical storm in your haste to submit your answers fast. This figure explains the correct answer.



VIEW - PUZZLE 3 QUESTION



SOLUTION:

PUZZLE 3 - THE CUBICAL STORM

*CLICK ON THE IMAGE TO ENLARGE IT !!!

Your Answer should be sent in the following format :

To : mathsntech.nitdgp@gmail.com

Subject : Puzzle -3 Answer

(Body of the message )

Answer : ....

Solution : (Explaining how you reached at the solution)

Name :

Year :

Branch :

College :

Happy solving !!!

PUZZLE -2 SOLUTION

SOLUTION :

The ranks of WINDOW and LUGISO are 271, 338.

So if the ranks are to be in A.P then the ranks should be either 204, or 405.

The ranks of

CRAVED - 198

PIERAT - 417

POLRAS - 417

RICSAT - 417

PHRASE - 422
So none of these things could be the answer !!!

PUZZLE - 2 MARINO'S PECULIAR HABIT

CLICK ON THE IMAGE TO ENLARGE IT !!!





Your Answer should be sent in the following format :
To : mathsntech.nitdgp@gmail.com
Subject : Puzzle -2 Answer
(Body of the message )
Answer : ....
Solution : (Explaining how you reached at the solution)
Name :
Year :
Branch :
College :
Happy solving !!!

PUZZLE -1 Thompson's number Solution

Solution :

First concentrating on the statements 3,5,7. If statement 3 is to be satisfied, then 5 and 7 cannot be satisfied. If we consider the statement 3 to be true then the digits of the number must form a fibonacci series : 0112358. But since there are no constraints on the length of the numbr and the last two digits are not squares we eliminate 8. Therfore we have 011235 now. 5 cannot be included because the third digit is not the difference of first and second. We have 01123. Since there are two adjacent one's, zero cannot be included. Thus we are left with 1123, which satisfies all other conditions.

Solution Submitted by Arun Saragadam, IIT Kharagpur :

(Appreciating the effort put in to check all the cases and arrive at a single solution)

Lets assume 4 is true.Then the possible cases are:->Last three digits are 222,333,555,777,235,237,357 etcBut all 333 or 444 or 555 or 777 cant exist as the rules described by them will not let them exist. Similarly no two of them can coexist as one rule contradicts the other.Even 222 cant occur or atleast the number cant end in 2 as the 2 states the number is not even. Hence the possibilities are 223,233 etcEven this is not possible as their rules are against these.Hence, 4 is eliminated.

Lets assume now 0 is true.Hence, no two adjacent digits are equal which denies the existence of 0 along with 3,5,7 as according to their rules atleast two digits must be adjacent if 0 is one of the digits.Hence, 3,5,7 are eliminatedlets assume 1 is presentbut it says atleast one odd prime must be present Hence 1 is also eliminatedLets assume 2 is present which says the number is odd which requires the presence of 1,3,5,7 which eliminates 2 too.Lets assume 8 is present which says last two numbers are squares. hence they must be 09,90(00 and 99 are not possible as adjacent numbers must not be equal). But if 9 is present last two numbers must be even. Hence no other possibility exists. Hence 8 is also elimiinated.

Lets assume 9 is present.which says last two numbers must be even but we are left with no possibilities which eliminates 9 too.As 0 cant exist independently, this case is eliminatedNow, 0 and 4 are eliminated and we are left with 1,2,3,4,5,6,7,8,9

Lets assume 8 is presentLast two numbers can be therefore 11, 19, 99,91But whenever 9 is present it requires last two numbers to be even which leaves us with the only possibility of 11But 1 demands the presence of at least one of 3,5,7Suppose 3 is present 11 must be preceeded by 0 in order to satisfy rule no 3 hence 3 is eliminatedSuppose 5 is present 11 must be preceeded by 2 in order to satisfy rule no 5 and 3 before 2 (in order to reach 8) but 3 and 5 cant coexist hence 5 is eliminated too.Similarly the case of 7 is eliminated.Hence 8 is eliminated.
So, now 0,4,8 are eliminated and we are left with 1,2,3,5,6,7,9

Lets assume 9 is present Hence the last two digits can be 2 and 6 but if 2 is present the number cant be even which contradicts the possibility of 9. Hence 9 is eliminated.
So, now 1,4,8,9 are eliminated and we are left with 1,2,3,5,6,7

Suppose 6 is presentThere must be only one digit which is prime(2,3,5,7)If its 2 , the number cant be even and demands an odd number which must be 1 as other prime cant exit but 1 itself demands atleast one odd prime number which thus eliminates the possibility of 2.Suppose 3 or 5 or 7 is presentthere must exist another number which is nothing but 1 (as otherwise if only two numbers exist it means all the three rules 3,5,7 are true as there will be only two numbers and demands all the three which contradicts the rules)So, if 1 is present6 must be at the end as the number must be even otherwise 2 needs to be presentand there must be 6,1 and one of 3,5,6 as according to rule 6 there must be only one prime numbersuppose 3 exists3 and 1 cant add upto 6 hence 3 is eliminated5 -1 != 6 hence 5 is eliminated7-1 =6 but atleast one number must be present twice in order to prevent 0 from appearing but which is impossible as to satisfy rule 7hence 6 is eliminatedNow, we are left with 1,2,3,5,7
suppose 5 is presentaccording to 5 difference of two adjacent digits gives the third digit but here atleast one digit is to be present twice which is impossible with satisfying rule 5similarly the case with 7
Hence, 5,7 are eliminated

So, now we are left with 1,2,3Suppose 2 is presentthere should be another odd number present as the number cant be even according to rule no 2So, it should be 1 or 3 either case demands the third numberso all the three numbers must be presenthence in order to satisfy rule no 3 only the following number can be formed1123Hence, the answer is 1123

PUZZLE MANIA RESULTS !!!

PUZZLE 6 RESULTS :
The list of honours is shared by :
*click to enlarge (Only getting both the puzzles right is considered as CORRECT ENTRY)WINNER OF PUZZLE 6- Akash Jajoo, 3rd year, IT, NIT Durgapur

PUZZLE 5 RESULTS :

We recieved quite a few answers for the puzzle -5 but I am sorry to declare none of them could be deemed right. All the best for PUZZLE -6


PUZZLE 4 RESULTS :
The list of honours is shared by :
*click to enlarge
WINNER OF PUZZLE 4- Sayanta Sundar Bhar, 2nd year, CSE, NIT Durgapur



PUZZLE 3 RESULTS :
The list of honours is shared by :
*click to enlarge
WINNER OF PUZZLE 3- Saikat Choudury, 4th year, CSE, NIT Durgapur

PUZZLE 2 RESULTS :
The list of honours is shared by :
*click to enlarge

WINNER OF PUZZLE 2 - ANURAN CHATTARAJ , 4th year, ECE, NIT Durgapur

PUZZLE 1 RESULTS :

Winner of PUZZLE - 1 : Sayanta Sundar Bhar , CSE, 2nd Year, NIT Durgapur.
Best Solution given by : Arun Saragadam, CSE, 3rd Year, IIT Kharagpur.



Return to home page !!!

Puzzle 1 - Thompson's Number

CLICK ON THE IMAGE TO ENLARGE IT !!!

Your Answer should be sent in the following format :

To : mathsntech.nitdgp@gmail.com

Subject : Puzzle -1 Answer

(Body of the message )

Answer : ....

Solution : (Explaining how you reached at the solution)

Name :

Year :

Branch :

College :

Happy solving

SAMPLE PUZZLE

SAMPLE PUZZLE _ SOLUTION !!!

PROBLEM STATEMENT:
The dragon of ignorance has three heads and three tails. However, you can slay it with the sword of knowledge by cutting off all its heads and tails. With one swipe of the sword you can cut off one head, two heads, one tail, or two tails.
But . . .
When you cut off one head, a new one grows in its place.
When you cut off one tail, two new tails replace it.
When you cut off two tails, one new head grows.
When you chop off two heads, nothing grows.

Help the world by slaying the dragon of ignorance. What are the minimum number of CUTS required to kill the dragon?

SOLUTION :

All of the tails must be converted to heads in such a way that the dragon is left with an even number of heads. The way to do this in the smallest number of strokes would be to first convert the 3 tails to 6 tails by using 3 1-tail strokes, then converting the 6 tails to 3 heads via 3 2-tail strokes. Finally, the now 6-headed dragon can be killed with 3 2-head strokes, for a total of 9 strokes.