Categories

## 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.

Categories

## 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.

Categories

## 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.

Categories

## Area of Equilateral Triangle | AIME I, 2015 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry, based on Area of Equilateral Triangle (Question 4).

## Area of Triangle – AIME I, 2015

Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find $x^{2}$.

• is 107
• is 507
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Theory of Equations

Geometry

AIME, 2015, Question 4

Geometry Revisited by Coxeter

## Try with Hints

Let A(0,0), B(16,0),C(20,0). let D and E be in first quadrant. then D =$(8,8\sqrt3)$, E=$(18,2\sqrt3$).

M=$(9,\sqrt3)$, N=($14,4\sqrt3$), where M and N are midpoints

since BM, BN, MN are all distance, BM=BN=MN=$2\sqrt13$. Then, by area of equilateral triangle, x=$13\sqrt3$ then$x^{2}$=507.

Categories

## Probability Problem | Combinatorics | AIME I, 2015 – Question 5

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

## Probability Problem – AIME I, 2015

In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 341
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Theory of Equations

Probability

AIME, 2015, Question 5

Geometry Revisited by Coxeter

## Try with Hints

Wednesday case – with restriction , select the pair on wednesday in $5 \choose 1$ ways

Tuesday case – four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is $8 \choose 2$ – 4

Monday case – a total of 6 socks and a pair not picked $6 \choose 2$ -2

by multiplication and principle of combinatorics $\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}$=$\frac{26}{315}$. That is 341.

Categories

## Exponents and Equations | AIME I, 2010 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Exponents and Equations.

## Exponents and Equations – AIME 2010

Suppose that y=$\frac{3x}{4}$ and $x^{y}=y^{x}$. The quantity x+y can be expressed as a rational number $\frac{r}{s}$ , where r and s are relatively prime positive integers. Find r+s.

.

• is 107
• is 529
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Number Theory

AIME, 2010, Question 3.

Elementary Number Theory by Sierpinsky

## Try with Hints

y=$\frac{3x}{4}$ into  $x^{y}=y^{x}$  and $x^{\frac{3x}{4}}$=$(\frac{3x}{4})^{x}$ implies $x^{\frac{3x}{4}}$=$(\frac{3}{4})^{x}x^{x}$ implies $x^{-x}{4}$=$(\frac{3}{4})^{x}$ implies $x^{\frac{-1}{4}}=\frac{3}{4}$ implies $x=\frac{256}{81}$.

y=$\frac{3x}{4}=\frac{192}{81}$.

x+y=$\frac{448}{81}$ then 448+81=529.

Categories

## Theory of Equations | AIME I, 2015 | Question 10

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

## Theory of Equations – AIME 2015

Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

• is 107
• is 72
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Algebra

AIME, 2015, Question 10.

Polynomials by Barbeau.

## Try with Hints

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

Categories

## Cube of Positive Integer | Number Theory | AIME I, 2015 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

## Cube of Positive Numbers – AIME I, 2015

There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

• is 107
• is 183
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Theory of Equations

Number Theory

AIME, 2015, Question 3

Elementary Number Theory by David Burton

## Try with Hints

$a^{3}=12p+1$ implies that $a^{3}-1=12p$ that is (a-1)($a^{2}$+a+1)=12p

a is odd, a-1 even, \(a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13

\(a^{2}+a+1=p implies p= 169+13+1=183.