Unlocking Math Olympiad Success: Navigating the Maze of Fake vs. Real Olympiads

Hey there, math enthusiasts and aspiring Olympiad champions! 🌟

Welcome back to our Math-related discussion, where we dive deep into the enchanting world of mathematics. Today, we're going to embark on a journey that's both exciting and crucial for every young math prodigy - the quest from Fake Math Olympiads to Real Math Olympiads!

The Olympiad Odyssey Begins

Picture this: Thousands and thousands of bright-eyed students in India, full of mathematical dreams, gearing up for the first level of the prestigious Indian Mathematical Olympiad (IOQM). They're armed with certificates, silver medals, and even gold medals from private Olympiads in their younger years, brimming with excitement and anticipation.

But then, reality strikes, and it's not pretty. Many of these talented youngsters are met with single-digit scores, and some even land a big, fat zero. It's disheartening, to say the least. These students, who are brimming with potential and love for mathematics, find their dreams crashing like a house of cards.

The Fake Olympiad Trap

At Cheenta, our mission is to nurture young mathematical minds, and we've seen this pattern too many times. We've got around five or six hundred students with us right now, and we've been preaching the same message over and over again: "Beware of fake Olympiads!" Why, you ask? Well, here's the scoop.

1. Question Quality: The questions in these private Olympiads often don't even come close to the real deal. They're not challenging enough and don't adequately prepare you for the IOQM.

2. False Hopes: These fake Olympiads dangle the tantalizing carrot of medals and certificates, making students believe they're math geniuses. But when the IOQM reality check arrives, it's often a rude awakening.

Real Olympiads vs. Fake Ones

Now, you might wonder, how do you tell the real Olympiads from the fakes? It's simpler than you think.

1. Organizer Matters: Genuine Olympiads are organized by reputable mathematical associations like the Mathematical Association of America or the Association of Mathematics Teachers of India (AMTI). These contests are gold mines of challenging problems crafted by true mathematical wizards.

2. Avoid Money-Making Schemes: Private Olympiads are often designed solely to make money, and they thrive on false promises. Those shiny gold medals? They're more about marketing than merit.

Nurturing Olympiad Success

So, how did our Cheenta students manage to shine in the IOQM this year? The secret is simple, but it requires dedication. We focus on problem-solving workshops, dedicating five days a week to Olympiad mathematics. It's all about non-routine mathematics and nothing else.

Your Path to Olympiad Glory

Now, here's the action plan for you, whether you're with Cheenta or charting your own course:

1. Consistency is Key: Dedicate at least one hour every day to non-routine mathematics. Consistency is the magic ingredient that fuels cumulative progress.

2. Aim for 3,000 Problems: Challenge yourself to tackle 3,000 problems in a year. For our students, that's 1,000 in class and 2,000 as homework. It's a starting point, but a strong one.

Embrace the Math Journey

Remember, mathematics is not just a subject; it's a world of wonder waiting to be explored. I'm still immersed in it, even at my age, because it's an endless adventure. Dive into the world of mathematics, seek its beauty, and you'll find yourself enchanted.

Join the Conversation!

This might be a one-sided conversation, but it doesn't have to be. Share your thoughts in the comments. What are your experiences with fake Olympiads? Let's discuss and learn together. 🌟

Watch the video by Dr. Ashani Dasgupta
Founder - faculty at Cheenta.

What are opportunities after Math Olympiad?

Watch the video to learn more about opportunities after Mathematical Olympiads in India, the United States and other countries.

A beautiful book from Eastern Europe for Math Olympiads, ISI CMI Entrance and joy of doing math

Selected Problems and Theorems in Elementary Mathematics – Arithmetic and Algebra by D. O. Shklyarsky,  N. N. Chentsov and I. M. Yaglom. T

This book contains The conditions of problems, the answers and hints  to them and the solutions of the problems. The conditions of the most difficult problems are marked by stars. We recommend the reader to start with trying to solve without assistance the problem he is interested in. In case this attempt
fails he can read the hint or the answer to the problem, which may facilitate the solution, Finally, if this does not help, the solution of the problem given in the book should be studied. However, for the starred problems it may turn out to be appropriate to begin with reading the hints or the answers before proceeding to solve the problems.

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Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations - AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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Theory of Equations | AIME I, 2015 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.

Theory of Equations - AIME I, 2015


The expressions A=\(1\times2+3\times4+5\times6+...+37\times38+39\)and B=\(1+2\times3+4\times5+...+36\times37+38\times39\) are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.

  • is 722
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Equations

Number Theory

Check the Answer


Answer: is 722.

AIME I, 2015, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


A = \((1\times2)+(3\times4)\)

\(+(5\times6)+...+(35\times36)+(37\times38)+39\)

B=\(1+(2\times3)+(4\times5)\)

\(+(6\times7)+...+(36\times37)+(38\times39)\)

B-A=\(-38+(2\times2)+(2\times4)\)

\(+(2\times6)+...+(2\times36)+(2\times38)\)

=722.

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Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

Probability in Games - AIME I, 1999 Question 13


Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)

  • 10
  • 742
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Combinations

Check the Answer


Answer: 742.

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

Try with Hints


\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes

no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)

the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.

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Equations with number of variables | AIME I, 2009 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.

Equations with number of variables - AIME 2009


For t=1,2,3,4, define \(S^{t}=a^{t}_1+a^{t}_2+...+a^{t}_{350}\), where \(a_{i}\in\){1,2,3,4}. If \(S_{1}=513, S_{4}=4745\), find the minimum possible value for \(S_{2}\).

  • is 905
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Number Theory

Check the Answer


Answer: is 905.

AIME, 2009, Question 14

Polynomials by Barbeau

Try with Hints


j=1,2,3,4, let \(m_{j}\) number of \(a_{i}\) s = j then \(m_{1}+m{2}+m{3}+m{4}=350\), \(S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513\) \(S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745\)

Subtracting first from second, then first from third yields \(m_{2}+2m_{3}+3m_{4}=163,\) and \(15m_{2}+80m_{3}+255m_{4}=4395\) Now subtracting 15 times first from second gives \(50m_{3}+210m_{4}=1950\) or \(5m_{3}+21m_{4}=195\) Then \(m_{4}\) multiple of 5, \(m_{4}\) either 0 or 5

If \(m_{4}=0\) then \(m_{j}\) s (226,85,39,0) and if \(m_{4}\)=5 then \(m_{j}\) s (215,112,18,5) Then \(S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917\) and \(S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905\) Then min 905.

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Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin - AIME I, 2009 Question 3


A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.

  • 10
  • 20
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Polynomials

Check the Answer


Answer: 11.

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints


here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)

then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)

then p=\(\frac{5}{6}\) then m+n=11

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Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

Geometric Sequence Problem - AIME 2009


Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

  • is 500
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Series

Real Analysis

Check the Answer


Answer: is 840.

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

Try with Hints


3-digit sequence a, ar, \(ar^{2}\). The largest geometric number must have a<=9.

ar \(ar^{2}\) less than 9 r fraction less than 1 For a=9 is \(\frac{2}{3}\) then number 964.

a>=1 ar and \(ar^{2}\) greater than 1 r is 2 and number is 124. Then difference 964-124=840.

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Inequality | HANOI 2018

Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Inequality.

Inequality - HANOI 2018


Find the number of integers that satisfy the inequality \(n^{4}-n^{3}-3n^{2}-3n-17 \lt 0\).

  • is 4
  • is 6
  • is 8
  • cannot be determined from the given information

Key Concepts


Algebra

Theory of Equations

Inequality

Check the Answer


Answer: is 4.

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

Try with Hints


We have \((n+1)^{3}+16 \gt n^{4} \geq 0\) which implies \(n \geq -3\).

For \(n \geq 4\) we have \(n^{4}-(n+1)^{3}\) \(\geq 3n^{3}-3n^{2}-3n-1\) \(\geq 12n^{2}-3n^{2}-3n-1\) \(=n(n-3)+8n^{2}-1 \gt 16\).

Then \(-3 \leq n \leq 3\). By directly calculation we obtain n=-1,0,1,2 that is 4 such integers.

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