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Algebra Math Olympiad PRMO USA Math Olympiad

Value of Sum | PRMO – 2018 | Question 16

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

Value of Sum – PRMO 2018, Question 16


What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

  • $50$
  • $53$
  • $55$
  • $59$
  • $65$

Key Concepts


Odd-Even

Sum

integer

Check the Answer


But try the problem first…

Answer:$55$

Source
Suggested Reading

PRMO-2018, Problem 16

Pre College Mathematics

Try with Hints


First hint

We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……

Can you now finish the problem ……….

Second Hint

$i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$
$=1-4+10=7$
$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$

$i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$
$=3-3+10=10$
$i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10$
$=13$
$i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16$
$i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$ 1 \times 9+(10)=19$

Can you finish the problem……..

Final Step

Therefore $ S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55 $



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Algebra Math Olympiad PRMO USA Math Olympiad

Digits Problem | PRMO – 2018 | Question 19

Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.

Digits Problem – PRMO 2018, Question 19


Let $N=6+66+666+\ldots \ldots+666 \ldots .66,$ where there are hundred 6 ‘s in the last term in the sum. How many times does the digit 7 occur in the number $N ?$

  • $30$
  • $33$
  • $36$
  • $39$
  • $42$

Key Concepts


Number theorm

Digits Problem

integer

Check the Answer


But try the problem first…

Answer:$33$

Source
Suggested Reading

PRMO-2018, Problem 19

Pre College Mathematics

Try with Hints


First hint

Given that $\mathrm{N}=6+66+666+……. \underbrace{6666 …..66}_{100 \text { times }}$

If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply \(\frac{6}{9}\) then it becomes $=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$

Can you now finish the problem ……….

Second Hint

$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right]$

Can you finish the problem……..

Final Step

$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60$

$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times



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Categories
Algebra AMC 10 Coordinate Geometry Math Olympiad USA Math Olympiad

Problem on Curve | AMC 10A, 2018 | Problem 21

Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Curve- AMC 10A, 2018- Problem 21


Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?

  • $a=\frac{1}{4}$
  • $\frac{1}{4}<a<\frac{1}{2}$
  • $a>\frac{1}{4}$
  • $a=\frac{1}{2}$
  • $a>\frac{1}{2}$

Key Concepts


Algebra

greatest integer

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-14

Check the answer here, but try the problem first

$a>\frac{1}{2}$

Try with Hints


First Hint

We have to find out the value of \(a\)

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and \(a\). if you solve this equation you will get the value of \(a\)

Now can you finish the problem?

Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either \(x^2=0\Rightarrow x=0\) or \(x^2-(2a-1)=0\)

\(\Rightarrow x=\pm \sqrt {2a-1}\). Therefore

Now Can you finish the Problem?

Third Hint

Therefore \(\sqrt {2a-1} > 0\)

\(\Rightarrow a>\frac{1}{2}\)

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Algebra AMC 10 Math Olympiad USA Math Olympiad

Finding Greatest Integer | AMC 10A, 2018 | Problem No 14

Try this beautiful Problem on Algebra based on finding greatest integer from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Finding Greatest Integer – AMC-10A, 2018- Problem 14


What is the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

  • $80$
  • $81$
  • $96$
  • $97$
  • $625$

Key Concepts


Algebra

greatest integer

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-14

Check the answer here, but try the problem first

$80$

Try with Hints


First Hint

The given expression is $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

We have to find out the greatest integer which is less than or equal to the given expression .

Let us assaume that $x=3^{96}$ and $y=2^{96}$

Therefore the given expression becoms $\frac{81 x+16 y}{x+y}$

Now can you finish the problem?

Second Hint

Now $\frac{81 x+16 y}{x+y}$

=$\frac{16 x+16 y}{x+y}+\frac{65 x}{x+y}$

$=16+\frac{65 x}{x+y}$

Now if we look very carefully we see that $\frac{65 x}{x+y}<\frac{65 x}{x}=65$

Therefore $16+\frac{65 x}{x+y}<16+65=81$

Now Can you finish the Problem?

Third Hint

Therefore less than \(81\) , the answer will be \(80\)

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Algebra Math Olympiad PRMO

Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Sum of Sides of Triangle from Triangle from Prmo-2018, Problem 17.

Sum of Sides of Triangle – PRMO, 2018- Problem 17


Triangles $\mathrm{ABC}$ and $\mathrm{DEF}$ are such that $\angle \mathrm{A}=\angle \mathrm{D}, \mathrm{AB}=\mathrm{DE}=17, \mathrm{BC}=\mathrm{EF}=10$ and $\mathrm{AC}-\mathrm{DF}=12$
What is $\mathrm{AC}+$ DF ?

,

  • \(28\)
  • \(30\)
  • \(21\)
  • \(26\)
  • \(26\)

Key Concepts


Geometry

Triangle

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

Prmo-2018, Problem-17

Check the answer here, but try the problem first

\(30\)

Try with Hints


First Hint

In \(\triangle ABC\) & \(\triangle DEF\), given that \(\angle A\)=\(\angle D\) & \(AB=DE\)=$17$ . $BC = EF = 10 $ and $AC – DF = 12.$ we have to find out $AC+DF$.

According to the question Let us assume that the point A coincides with D, B coincides with E. Now if we draw a circle with radius 10 and E(B) as center ….

Can you finish the problem?

Second Hint

Let M be the foot of perpendicular from B(E) to CF. So BM = 8. Hence $\mathrm{AM}=\sqrt{17^{2}-8^{2}}=\sqrt{(25)(9)}=15$

Third Hint

Hence $AF = 15 – 6 = 9$ & $AC = 15 + 6 = 21$

So $AC + DF = 30$

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Categories
Algebra AMC 10 Math Olympiad USA Math Olympiad

Linear Equation Problem | AMC 10A, 2015 | Problem No.16

Try this beautiful Problem on Algebra from the Linear equation from AMC 10 A, 2015.

Linear Equation Problem – AMC-10A, 2015- Problem 16


If $y+4=(x-2)^{2}, x+4=(y-2)^{2}$, and $x \neq y$, what is the value of $x^{2}+y^{2} ?$

,

  • $11$
  • $12$
  • $15$
  • $14$
  • $6$

Key Concepts


Algebra

Equation

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2015 Problem-16

Check the answer here, but try the problem first

$15$

Try with Hints


First Hint

Given that $y+4=(x-2)^{2}, x+4=(y-2)^{2}$ . we have to find out $x^{2}+y^{2} ?$. Now add two equations $x^{2}+y^{2}-4 x-4 y+8=x+y+8$

\(\Rightarrow x^{2}+y^{2}=5(x+y)\)

Can you find out the value \(x+y\)?

Second Hint

We can also subtract the two equations to yield the equation
$x^{2}-y^{2}-4 x+4 y=y-x$

\(\Rightarrow x^{2}-y^{2}=(x+y)(x-y)=3 x-3 y=3(x-y)\)

Therefore \((x+y)(x-y)=3 x-3 y=3(x-y)\)

\(\frac{(x+y)(x-y)}{(x-y)}=\frac{3(x-y)}{(x-y)}\) [ as\( x \neq y\)]

\(\Rightarrow (x+y)=3\)

Third Hint

Therefore \(x^2+y^2=5(x+y)=5 \times 3=15\)

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Categories
Algebra AMC 10 Math Olympiad USA Math Olympiad

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

Contents
 [hide]

    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


    But try the problem first…

    Answer: \((2018)\)

    Source
    Suggested Reading

    AMC-10A (2019) Problem 19

    Pre College Mathematics

    Try with Hints


    First hint

    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 ……….

    Second Hint

    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……..

    Final Step

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

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    Categories
    Algebra Arithmetic Math Olympiad PRMO

    Nearest value | PRMO 2018 | Question 14

    Try this beautiful problem from the PRMO, 2018 based on Nearest value.

    Nearest Value – PRMO 2018


    If x=cos1cos2cos3…..cos89 and y=cos2cos6cos10….cos86, then what is the integer nearest to \(\frac{2}{7}log_2{\frac{y}{x}}\)?

    • is 107
    • is 19
    • is 840
    • cannot be determined from the given information

    Key Concepts


    Algebra

    Numbers

    Multiples

    Check the Answer


    But try the problem first…

    Answer: is 19.

    Source
    Suggested Reading

    PRMO, 2018, Question 14

    Higher Algebra by Hall and Knight

    Try with Hints


    First hint

    \(\frac{y}{x}\)=\(\frac{cos2cos6cos10…..cos86}{cos1cos2cos3….cos89}\)

    =\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10…cos86}{sin2sin4…sin88}\)

    [ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]

    Second Hint

    =\(\frac{2^{\frac{89}{2}}sin4sin8sin12…sin88}{sin2sin4sin6…sin88}\)

    [ since sin\(\theta\)=cos(90-\(\theta\))]

    =\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)

    [ since cos\(\theta\)=sin(90-\(\theta\))]

    Final Step

    =\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)

    [since \(cos4cos8cos12…cos88\)

    \(=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60\)

    \(=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)\)

    \(=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)\)

    \(=(1/2)^{20}(cos36cos72)\)

    \(=(1/2)^{20}(cos36sin18)\)

    \(=(1/2)^{22}(4sin18cos18cos36/cos18)\)

    \(=(1/2)^{22}(sin72/cos18)\)

    \(=(1/2)^{22}\)]

    =\(2^\frac{133}{2}\)

    \(\frac{2}{7}log_2{\frac{y}{x}}\)=\(\frac{2}{7} \times \frac{133}{2}\)=19.

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    Categories
    AIME II Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

    Sequence and permutations | AIME II, 2015 | Question 10

    Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

    Sequence and permutations – AIME II, 2015


    Call a permutation \(a_1,a_2,….,a_n\) of the integers 1,2,…,n quasi increasing if \(a_k \leq a_{k+1} +2\) for each \(1 \leq k \leq n-1\), find the number of quasi increasing permutations of the integers 1,2,….,7.

    • is 107
    • is 486
    • is 840
    • cannot be determined from the given information

    Key Concepts


    Sequence

    Permutations

    Integers

    Check the Answer


    But try the problem first…

    Answer: is 486.

    Source
    Suggested Reading

    AIME II, 2015, Question 10

    Elementary Number Theory by David Burton

    Try with Hints


    First hint

    While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

    Second Hint

    Number of permutations with n elements is three times the number of permutations with n-1 elements

    or, number of permutations for n elements=3 \(\times\) number of permutations of (n-1) elements

    or, number of permutations for n elements=\(3^{2}\) number of permutations of (n-2) elements

    ……

    or, number of permutations for n elements=\(3^{n-2}\) number of permutations of {n-(n-2)} elements

    or, number of permutations for n elements=2 \(\times\) \(3^{n-2}\)

    forming recurrence relation as the number of permutations =2 \(\times\) \(3^{n-2}\)

    for n=3 all six permutations taken and go up 18, 54, 162, 486

    Final Step

    for n=7, here \(2 \times 3^{5} =486.\)

    Header text

    as

    Header text

    sds

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    Categories
    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


    But try the problem first…

    Answer: is 40.

    Source
    Suggested Reading

    AIME, 2012, Question 1.

    Elementary Number Theory by David Burton .

    Try with Hints


    First hint

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

    Second Hint

    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

    Final Step

    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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