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Arithmetic Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 997.

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.

.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 173.

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

taking k=1,2,….,100 and adding

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

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

\(\Rightarrow S=\frac{2525}{49}\)

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

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=75+98

=173.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 248.

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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AMC 10 Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


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=\(21\)

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 25.

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.

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AMC 10 Math Olympiad USA Math Olympiad

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

Check the Answer


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 \(78\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


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 \(1\)

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Algebra Math Olympiad PRMO USA Math Olympiad

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

Check the Answer


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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Algebra Math Olympiad PRMO USA Math Olympiad

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

Check the Answer


Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

Try with Hints


Rectangle Problem

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 ……….

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

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 870.

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.

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