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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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Algebra Arithmetic Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

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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Algebra Arithmetic Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

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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Algebra Arithmetic Functional Equations Math Olympiad Math Olympiad Videos USA Math Olympiad

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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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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AIME I Complex Numbers Math Olympiad Math Olympiad Videos USA Math Olympiad

Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers – AIME, 2009


There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

  • 101
  • 201
  • 301
  • 697

Key Concepts


Complex Numbers

Theory of equations

Polynomials

Check the Answer


Answer: 697.

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints


Taking z=a+bi

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

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AIME I Algebra Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009


A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

  • 110
  • 420
  • 430
  • 111

Key Concepts


Combinations

Theory of equations

Polynomials

Check the Answer


Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints


Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

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AMC 10 Math Olympiad Videos

Extremal Principle for Counting – AMC 10

Extremal Principle is used in a variety of problems in Math Olympiad. The following problem from AMC 10 is a very nice example of this idea.

AMC 10 Problem 4 (2019)- Based on Extremal Principle

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

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Try this problem. Send answers to helpdesk@cheenta.com

Suppose there 5 blue balls, 7 white balls, and 9 green balls. At least how many balls should you pick up (without looking a and without replacement) to be sure that you have picked up at least 4 balls of the same color?

You can also try these problems related to spiral similarity.

You may also click the link to learn its application:- https://www.youtube.com/watch?v=8o8AAWt960o

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AMC 8 Math Olympiad Videos USA Math Olympiad Videos

AMC 8 2019 – Stick and Dot Method

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Understand the problem

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[/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0.9″ open=”off”]CombinatoricsĀ 

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7/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of MathematicsĀ 

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Watch video

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