M Decoder

Are you witty enough to beat 'M-Coders'?

Imagine yourself lost in a maze full of challenges. The only way to get out is by using your problem-solving and analytical skills. Presenting a range of puzzles and problems to solve, this one-of-a-kind    mathematical event seeks to bring out the best candidate amongst many.

Are you the invincible candidate? Test your skills in this intriguing battle of wits.

Event starts at 11:59 pm, February 27

CASH PRIZE - INR 15000

It is a 6 day online event. Everyday, 6 new questions with different weights and an encrypted puzzle will be presented. The event will happen during 27th February to 3rd March 2017.

The participant has to solve only 1 out of the 6 questions (mandatory) everyday and he may solve the encrypted puzzle (not mandatory) for bonus points. Now how should a participant select 1 question from the pool of 6 questions to earn maximum points?

Solving the encrypted  puzzle initially (which gives a sequence of numbers) will help the participant get an idea on the weights associated with all the 6 questions presented that day.

The participant will then have to choose a question number which he thinks will fetch him the maximum points. Once he locks a question number, that particular question alone will be displayed. He/She cannot go back to change the question number.

Note: Solving the question having a higher weight will increase the chances of getting maximum points.

To aid the participant to get a “better score”(check out Scoring) , encrypted information in

the form of a puzzle, an equation and a graph will be presented to the participant.

Event Equation: AX=B

A (6*6 matrix) - Given to participant (which is frequently updated by the server to maintain the equation AX=B).

X (6*1 matrix) - To be encrypted from puzzle.

B (6*1 matrix) - Gives information about the number of participants who have solved each question.

Graph: Sum of all participants who solved any problem vs Time

Correlate Matrix B and the graph in order to identify the number of people who have solved the problem and hence the most efficient scoring option at that particular time.

The Bonus answer will be completely related to the puzzle. i.e., the sum of all the values of matrix X is the bonus answer.

So, solving the puzzle not only gets the person bonus points but also helps him get Matrix

B(6*1 matrix), which hints at the number of participants who solved at that time.

The 6 questions will test basic concepts in algebra, combinatorics, geometry and related topics and common sense.

• The score (points) of the participant depends on the weightage of the question (more the

weightage, more the score) and also on the number of participants who have already solved  that question at that time (i.e. more the participants who have already solved that

question before you, lesser will be your score).

• Points will be awarded for solving the puzzle and the mandatory question from the pool. However, the participant will not be allowed to solve the puzzle once he chooses a question from the pool of 6 questions.

• The scoring formula:

1. Score of the participant will be  S = 340 + (W * 10) - [(11- W) * (N -1)]   (in points).

This is only for the mandatory question. (from the pool of 6 questions)

where

W – Weightage of the question (1,2,3,4,5,6).

N – Number of people who have already solved that question before.

2. Bonus part (ie. encrypted puzzle) :  50 points.

This is an individual event. Only one question out of the 6 can be attempted per day.

1. What if the matrix is not decoded? Can we still solve the questions?

Ans. Yes, you can choose a random question number and solve it. But your chances of getting a high score will be limited. You cannot come back and solve the puzzle once you select the question number.

2. What is the most common strategy?

Ans. Let’s say after solving the puzzle, you see a question with high difficulty, which can potentially fetch you many points.

But suppose you find that many people have already solved this question, and only a few people have solved a relatively easier question. In this case, solving the easier question is probably more profitable, since the score of a participant also depends on the number of people who have previously solved the question. So the most common strategy is to decode the matrix and correlate it with the graph in order to get maximum number of points.

None whatsoever.

Coming up Soon!!!

Rishab          - 8122842729

Arvindh        - 8903607798

Sasi Kalyan  -9791271190

Email us at mdecoder@pragyan.org

Coming up Soon!!!