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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

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 $y_{1}+y_{2}=25$.

Categories

## 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

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

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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## 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

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

Categories

## 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

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

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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## 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

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

Categories

## 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

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

Categories

## 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?

## 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.

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# Understand the problem

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[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0.9″]American Mathematical Contest 2019, AMC 8 Problem 25

[/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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