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Algebra AMC 10 Math Olympiad USA Math Olympiad

Least Possible Value Problem | AMC-10A, 2019 | Quesstion19

Try this beautiful problem from Algebra based on Least Possible Value.

Least Possible Value – AMC-10A, 2019- Problem 19


What is the least possible value of \(((x+1)(x+2)(x+3)(x+4)+2019)\)

where (x) is a real number?

  • \((2024)\)
  • \((2018)\)
  • \((2020)\)

Key Concepts


Algebra

quadratic equation

least value

Check the Answer


Answer: \((2018)\)

AMC-10A (2019) Problem 19

Pre College Mathematics

Try with Hints


To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)

Let us take \(((x^2+5x+5=m))\)

then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)

Can you now finish the problem ……….

Clearly in \((m^2+2018)…….(m^2)\) is positive ( squares of any number is non-negative) and least value is 0

can you finish the problem……..

Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

Numbers of positive integers – AIME 2012


Find the number of positive integers with three not necessarily distinct digits, \(abc\), with \(a \neq 0\) and \(c \neq 0\) such that both \(abc\) and \(cba\) are multiples of \(4\).

  • is 107
  • is 40
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Number Theory

Algebra

Check the Answer


Answer: is 40.

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

Try with Hints


Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives \(10 \times 2 \times 2\)

Then number of ways \(10 \times 2 \times 2\) = 40 ways.

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Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

Arithmetic Sequence Problem – AIME 2012


The terms of an arithmetic sequence add to \(715\). The first term of the sequence is increased by \(1\), the second term is increased by \(3\), the third term is increased by \(5\), and in general, the \(k\)th term is increased by the \(k\)th odd positive integer. The terms of the new sequence add to \(836\). Find the sum of the first, last, and middle terms of the original sequence.

  • is 107
  • is 195
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Number Theory

Algebra

Check the Answer


Answer: is 195.

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

Try with Hints


After the adding of the odd numbers, the total of the sequence increases by \(836 – 715 = 121 = 11^2\).

Since the sum of the first \(n\) positive odd numbers is \(n^2\), there must be \(11\) terms in the sequence, so the mean of the sequence is \(\frac{715}{11} = 65\).

Since the first, last, and middle terms are centered around the mean, then \(65 \times 3 = 195\)

Hence option B correct.

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Geometry Math Olympiad USA Math Olympiad

Length and Triangle | AIME I, 1987 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

Length and Triangle – AIME I, 1987


Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA. Find PC.

Length and Triangle
  • is 107
  • is 33
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Triangles

Check the Answer


Answer: is 33.

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


Let PC be x, \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA=120 (in degrees)

Applying cosine law \(\Delta\)APB, \(\Delta\)BPC, \(\Delta\)CPA with cos120=\(\frac{-1}{2}\) gives

\(AB^{2}\)=36+100+60=196, \(BC^{2}\)=36+\(x^{2}\)+6x, \(CA^{2}\)=100+\(x^{2}\)+10x

By Pathagorus Theorem, \(AB^{2}+BC^{2}=CA^{2}\)

or, \(x^{2}\)+10x+100=\(x^{2}\)+6x+36+196

or, 4x=132

or, x=33.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

Algebra and Positive Integer – AIME I, 1987


What is the largest positive integer n for which there is a unique integer k such that \(\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}\)?

  • is 107
  • is 112
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 112.

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

Try with Hints


Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

for 91n-104k<n+k, K>\(\frac{6n}{7}\)

and 0<91n-104k gives k<\(\frac{7n}{8}\)

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Algebra Math Olympiad PRMO USA Math Olympiad

Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer – PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

  • $9$
  • $7$
  • $28$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

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Geometry Math Olympiad USA Math Olympiad

Distance and Spheres | AIME I, 1987 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres.

Distance and Sphere – AIME I, 1987


What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)?

  • is 107
  • is 137
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Spheres

Check the Answer


Answer: is 137.

AIME I, 1987, Question 2

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


The distance between the center of the spheres is \(\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}\)

=\(\sqrt{14^{2}+18^{2}+21^{2}}\)=31

The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.

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Categories
Algebra Arithmetic Math Olympiad USA Math Olympiad

Arithmetic Mean | AIME I, 2015 | Question 12

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

Arithmetic Mean of Number Theory – AIME 2015


Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

  • is 107
  • is 431
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequalities

Algebra

Number Theory

Check the Answer


Answer: is 431.

AIME, 2015, Question 12

Elementary Number Theory by David Burton

Try with Hints


Each 1000-element subset \({ a_1, a_2,a_3,…,a_{1000}}\) of \({1,2,3,…,2015}\) with \(a_1<a_2<a_3<…<a_{1000}\) contributes \(a_1\) to sum of least element of each subset and set \({a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\). \(a_1\) ways to choose a positive integer \(k\) such that \(k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1\) (\(k\) can be anything from \(1\) to \(a_1\) inclusive

Thus, the number of ways to choose the set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is equal to the sum. But choosing a set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is same as choosing a 1001-element subset from \({1,2,3,…,2016}\)!

average =\(\frac{2016}{1001}\)=\(\frac{288}{143}\). Then \(p+q=288+143={431}\)

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Algebra Combinatorics Math Olympiad USA Math Olympiad

Algebra and Combination | AIME I, 2000 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebra and Combination.

Algebra and combination – AIME 2000


In expansion \((ax+b)^{2000}\) where a and b are relatively prime positive integers the coefficient of \(x^{2}\) and \(x^{3}\) are equal, find a+b

  • is 107
  • is 667
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Combination

Check the Answer


Answer: is 667.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


 here coefficient of \(x^{2}\)= coefficient of \(x^{3}\) in the same expression

then \({2000 \choose 1998}a^{2}b^{1998}\)=\({2000 \choose 1997}a^{3}b^{1997}\)

then \(b=\frac{1998}{3}\)a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

.

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Algebra Functional Equations Math Olympiad USA Math Olympiad

Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

Algebraic Equation – AIME 2000


Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, \(x+\frac{1}{z}=5\) and \(y+\frac{1}{x}=29\) then \(z+\frac{1}{y}\)=\(\frac{m}{n}\) where m and n are relatively prime, find m+n

  • is 107
  • is 5
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

Try with Hints


 here \(x+\frac{1}{z}=5\) then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and \(y=(29-\frac{1}{x}\)) together gives 5-x=x\((29-\frac{1}{x}\)) then x=\(\frac{1}{5}\)

then y=29-5=24 and z=\(\frac{1}{5-x}\)=\(\frac{5}{24}\)

\(z+\frac{1}{y}\)=\(\frac{1}{4}\) then 1+4=5.

.

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