Categories

## Arithmetic and geometric mean | AIME I, 2000 Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

## Arithmetic and geometric mean with Algebra – AIME 2000

Find the number of ordered pairs (x,y) of integers is it true that $0 \lt y \lt 10^{6}$ and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

• is 107
• is 997
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Ordered pair

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

given that $\frac{x+y}{2}=2+({xy})^\frac{1}{2}$ then solving we have $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and-2

given that $y \gt x$ then $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and here maximum integer value of $y^\frac{1}{2}$=$10^{3}-1$=999 whose corresponding $x^\frac{1}{2}$=997 and decreases upto $y^\frac{1}{2}$=3 whose corresponding $x^\frac{1}{2}$=1

then number of pairs ($x^\frac{1}{2}$,$y^\frac{1}{2}$)=number of pairs of (x,y)=997.

.

Categories

## Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

## Sequence and fraction – AIME I, 2000

A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 173
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

Let S be the sum of the sequence $x_k$

given that $x_k=S-x_k-k$ for any k

$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$

$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$

$\Rightarrow S=\frac{2525}{49}$

for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$

$\Rightarrow x_{50}=\frac{75}{98}$

$\Rightarrow m+n$=75+98

=173.

Categories

## Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

## Sequence and greatest integer – AIME I, 2000

Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.

• is 107
• is 248
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

## Try with Hints

We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$

sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000

=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$

$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$

=2480 + $\frac{437}{1000}$

$\Rightarrow [\frac{s}{10}]$=248.

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## Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem.

## Arithmetic sequence – AMC-10A, 2015- Problem 7

How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$?

• $20$
• $21$
• $24$
• $60$
• $61$

### Key Concepts

Algebra

Arithmetic sequence

Answer: $21$

AMC-10A (2015) Problem 7

Pre College Mathematics

## Try with Hints

The given terms are $13$, $16$, $19$, $\dotsc$, $70$, $73$. We have to find out the numbers of terms…..

If you look very carefully then the distance between two digits is $3$.Therefore this is an Arithmetic Progression where the first term is $13$ and common difference is $3$

Can you now finish the problem ……….

$a+(n-1) d \Longrightarrow 13+(n-1) 3=73$

$\Rightarrow n=21$

The number of terms=$21$

Categories

## Cubic Equation | AMC-10A, 2010 | Problem 21

Try this beautiful problem based on Cubic Equation from AMC 10A, 2010.

## Cubic Equation – AMC-10A, 2010- Problem 21

The polynomial $x^{3}-a x^{2}+b x-2010$ has three positive integer roots. What is the smallest possible value of $a ?$

• $31$
• $78$
• $43$

### Key Concepts

Algebra

Cubic Equation

Roots

Answer: $78$

AMC-10A (2010) Problem 21

Pre College Mathematics

## Try with Hints

The given equation is $x^{3}-a x^{2}+b x-2010$

Comparing the equation with $Ax^3+Bx^2+Cx+D=0$ we get $A=1,B=-a,C=b,D=0$

Let us assume that $x_1,x_2,x_3$ are the roots of the above equation then using vieta’s formula we can say that $x_1.x_2.x_3=2010$

Therefore if we find out the factors of $2010$ then we can find out our requirement…..

can you finish the problem……..

$2010$ factors into $2 \cdot 3 \cdot 5 \cdot 67 .$ But, since there are only three roots to the polynomial,two of the four prime factors must be multiplied so that we are left with three roots and we have to find out the smallest positive values of $a$

can you finish the problem……..

To minimize $a, 2$ and 3 should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $78$

Categories

## LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

## Lcm and Integer – AIME I, 1998

Find the number of values of k in $12^{12}$ the lcm of the positive integers $6^{6}$, $8^{8}$ and k.

• is 107
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Lcm

Algebra

Integers

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

here $k=2^{a}3^{b}$ for integers a and b

$6^{6}=2^{6}3^{6}$

$8^{8}=2^{24}$

$12^{12}=2^{24}3^{12}$

lcm$(6^{6},8^{8})$=$2^{24}3^{6}$

$12^{12}=2^{24}3^{12}$=lcm of $(6^{6},8^{6})$ and k

=$(2^{24}3^{6},2^{a}3^{b})$

=$2^{max(24,a)}3^{max(6,b)}$

$\Rightarrow b=12, 0 \leq a \leq 24$

$\Rightarrow$ number of values of k=25.

Categories

## Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

## Pen & Note Books – PRMO 2017, Question 8

A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

• $9$
• $7$
• $11$

### Key Concepts

Algebra

Equation

multiplication

Answer:$7$

PRMO-2017, Problem 8

Pre College Mathematics

## Try with Hints

Given A pen costs Rs.$11$ and a note book costs Rs.$13$

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

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## Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

## Rectangle Problem – Geometry – PRMO 2017, Question 13

In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

• $9$
• $24$
• $11$

### Key Concepts

Geometry

Triangle

Trigonometry

Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

## Try with Hints

We have to find out the value of $AB$. Join $BD$. $BF$ is perpendicular on $AC$.

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Therefore

Therefore$\angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

Categories

## Problem on Fraction | AMC 10A, 2015 | Question 15

Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015.

## Fraction – AMC-10A, 2015- Problem 15

Consider the set of all fractions $\frac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by 1 , the value of the fraction is increased by $10 \%$ ?

,

• $0$
• $1$
• $2$
• $3$
• $4$

### Key Concepts

algebra

Fraction

Answer: $1$

AMC-10A (2015) Problem 15

Pre College Mathematics

## Try with Hints

Given that $\frac{x}{y},$ is a fraction where $x$ and $y$ are relatively prime positive integers. We have to find out the numbers of fraction if both numerator and denominator are increased by 1.

According to the question we have $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

Can you now finish the problem ……….

Now from the equation we can say that $x+1>\frac{11}{10} \cdot x$ so $x$ is at most 9
By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11 x}{10-x}$. since $x$ and $y$ are integer, this only has solutions for $x \in{5,8,9} .$ However, only the first yields a $y$ that is relative prime to $x$

can you finish the problem……..

Therefore the Possible answer will be $1$

Categories

## Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

## Integer – AIME I, 1993

Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

• is 107
• is 870
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Algebra

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

taking first two solutions a<b<c<d<500

or,$1 \leq c-93, c+1 \leq 499$

or, $94 \leq c \leq 498$ gives 405 solutions

and $1 \leq c-31, c+3 \leq 499$

or, $32 \leq c \leq 496$ gives 465 solutions

or, 405+465=870 solutions.