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

## Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

## Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of $2010^{2}$, she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

$(2+1)^{4}$ divisors, $2^{4}$ are squares

probability is $\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}$ implies m+n=107

Categories

## Coordinate Geometry Problem | AIME I, 2009 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

## Coordinate Geometry Problem – AIME 2009

Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

• is 107
• is 600
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Geometry

AIME, 2009, Question 11

Geometry Revisited by Coxeter

## Try with Hints

let P and Q be defined with coordinates; P=($x_1,y_1)$ and Q($x_2,y_2)$. Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=$\frac{1}{2}$OY.OX=$\frac{1}{2}$2009.49 And [OYP]=$\frac{1}{2}$$2009x_1$  and [OXQ]=$\frac{1}{2}$(49)$y_2$.

2009.49 is odd, area OYX not integer of form k+$\frac{1}{2}$ where k is an integer

41x+y=2009 taking both 25  $\frac{25!}{2!23!}+\frac{25!}{2!23!}$=300+300=600.

.

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

## Arrangement of digits | AIME I, 2012 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement of digits.

## Arrangement of digits – AIME 2012

Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

• is 107
• is 330
• is 840
• cannot be determined from the given information

### Key Concepts

Arrangements

Algebra

Number Theory

AIME, 2012, Question 5

Combinatorics by Brualdi

## Try with Hints

When 1 subtracts from a number, the number of digits remain constant when the initial number has units and tens place in 10

Then for subtraction from B requires one number with unit and tens place 10.

10 there, remaining 1 distribute any of other 11 then answer ${11 \choose 7} = {330}$.

Categories

## Complex Numbers and prime | AIME I, 2012 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and prime.

## Complex Numbers and primes – AIME 2012

The complex numbers z and w satisfy $z^{13} = w$ $w^{11} = z$ and the imaginary part of z is $\sin{\frac{m\pi}{n}}$, for relatively prime positive integers m and n with m<n. Find n.

• is 107
• is 71
• is 840
• cannot be determined from the given information

### Key Concepts

Complex Numbers

Algebra

Number Theory

AIME I, 2012, Question 6

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

Taking both given equations $(z^{13})^{11} = z$ gives $z^{143} = z$ Then $z^{142} = 1$

Then by De Moivre’s theorem, imaginary part of z will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}$ where $k \in {1, 2, upto 70}$

71 is prime and n = 71.

Categories

## Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

## Function and Symmetry – AIME I 1984

A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval $-1000 \leq x\leq 1000$.

• is 107
• is 401
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Symmetry

Number Theory

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

## Try with Hints

by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Then least number of roots as 401.

Categories

## Ratio of LCM & GCF | Algebra | AMC 8, 2013 | Problem 10

Try this beautiful problem from Algebra based on the ratio of LCM & GCF from AMC-8, 2013.

## Ratio of LCM & GCF | AMC-8, 2013 | Problem 10

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

• 310
• 330
• 360

### Key Concepts

Algebra

Ratio

LCM & GCF

Answer:$330$

AMC-8, 2013 problem 10

Challenges and Thrills in Pre College Mathematics

## Try with Hints

We have to find out the ratio of least common multiple and greatest common factor of 180 and 594. So at first, we have to find out prime factors of 180 & 594. Now…….

$180=3^2\times 5 \times 2^2$

$594=3^3 \times 11 \times 2$

Can you now finish the problem ……….

Now lcm of two numbers i.e multiplications of the greatest power of all the numbers

Therefore LCM of 180 & 594=$3^3\times2^2 \times 11 \times 5$=$5940$

For the GCF of 180 and 594, multiplications of the least power of all of the numbers i.e $3^2\times 2$=$18$

can you finish the problem……..

Therefore the ratio of Lcm & gcf of 180 and 594 =$\frac{5940}{18}$=$330$

Categories

## Unit digit | Algebra | AMC 8, 2014 | Problem 22

Try this beautiful problem from Algebra about unit digit from AMC-8, 2014.

## Unit digit | AMC-8, 2014|Problem 22

A $2$-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the unit digit of the number?

• 7
• 9
• 5

### Key Concepts

Algebra

Multiplication

integer

Answer:$9$

AMC-8, 2014 problem 22

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Let the ones digit place be y and ten’s place be x

Therefore the number be $10x+y$

Can you now finish the problem ……….

Given that the product of the digits plus the sum of the digits is equal to the number

can you finish the problem……..

Let the ones digit place be y and ten’s place be x

Therefore the number be $10x+y$

Now the product of the digits=$xy$

Given that the product of the digits plus the sum of the digits is equal to the number

Therefore $10x+y=(x\times y)+(x+y)$

$\Rightarrow 9x=xy$

$\Rightarrow y=9$

Therefore the unit digit =$y$=9

Categories

## Mixture | Algebra | AMC 8, 2002 | Problem 24

Try this beautiful problem from Algebra based on Mixture from AMC-8, 2002.

## Mixture | AMC-8, 2002 | Problem 24

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

• 34%
• 40%
• 26%

### Key Concepts

Algebra

Mixture

Percentage

AMC-8, 2002 problem 24

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Find the amount of juice that a pear and a orange can gives…

Can you now finish the problem ……….

Find total mixture

can you finish the problem……..

3 pear gives 8 ounces of juice .

A pear gives $\frac {8}{3}$ ounces of juice per pear

2 orange gives 8 ounces of juice per orange

An orange gives $\frac {8}{2}$=4 ounces of juice per orange.

Therefore the total mixer =${\frac{8}{3}+4}$

If She makes a pear-orange juice blend from an equal number of pears and oranges then percent of the blend is pear juice= $\frac{\frac{8}{3}}{\frac{8}{3}+4} \times 100 =40$