Problem based on Integer | PRMO-2018 | Problem 6

Try this beautiful problem from Algebra based on Integer

Algebra based on Integer PRMO Problem 6

Integers a, b, c satisfy $a + b – c = 1$ and $a^2 + b^2 – c^2 = –1$. What is the sum of all possible values of
$a^2 + b^2 + c^2$ ?

$24$
$18$
$34$

Key Concepts

Algebra

quadratic equation

Factorization

Check the Answer

Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

Try with Hints

Use $(a + b – 1 )= c$ this relation

Can you now finish the problem ..........

$(a + b – 1 )^2= c^2$ (squaring both sides.......)

Can you finish the problem........

Given that a,b,c are integer satisfy $a + b – c = 1$ and $a^2 + b^2 – c^2 = –1$.

Now

$(a + b – c )= 1$

$\Rightarrow (a + b – 1 )= c$

$\Rightarrow (a + b – 1 )^2= c^2$ (squarring both sides.......)

$\Rightarrow (a^2 + b^2 +1^2+2ab-2a-2b)= c^2$

$\Rightarrow (a^2 + b^2 –c^2+1+2ab)= 2(a+b)$

$\Rightarrow (-1)+1+2ab= 2(a+b)$ ( as $a^2 + b^2 – c^2 = –1$.

$\Rightarrow ab=a+b$

$\Rightarrow (a-1)(b-1)=1$

Therefore the possibile cases are $(a-1)=\pm 1$ and $(b-1)=\pm 1$

Therefore $ a=1$,$b=1$ or $a=0$,$b=0$

From the equation $(a + b – c )= 1$ ,C =3 (as a=b=2) and C=-1(as a=b=0)

Therefore $(a^2 +b^2 +c^2)=4+4+9=17$ and $(a^2 +b^2 +c^2)=0+0+1=1$

sum of all possible values of $a^2 + b^2 + c^2=17+1=18$

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?

Key Concepts

Algebra

Mixture

Percentage

Check the Answer

Answer:40%

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= $\frac{\frac{8}{3}}{\frac{8}{3}+4} \times 100 =40$

Quadratic Equation Problem | PRMO-2018 | Problem 9

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation PRMO Problem 9

Suppose a,b are integers and a + b is a root of $x^2 +ax+b=0$.What is the maximum possible
values of $ b^2 $?

$49$
$81$
$64$

Key Concepts

Algebra

quadratic equation

Factorization

Check the Answer

Answer:$81$

PRMO-2018, Problem 9

Pre College Mathematics

Try with Hints

(‘a+b”) is the root of the equation therefore (“a+b”) must satisfy the given equation

Can you now finish the problem ..........

Discriminant is a perfect square

can you finish the problem........

Given that $‘a+b”$ is the root of the equation therefore $“a+b”$ must satisfy the given equation

Therefore the given equation becomes ……

$(a+b)^2 +a(a+b)+b=0$

$\Rightarrow a^2 +2ab+b^2+a^2+ab+b=0$

$\Rightarrow 2a^2 +3ab+b^2+b=0$

Now since “a” is an integer,Discriminant is a perfect square

$\Rightarrow 9b^2 -8(b^2+b)=m^2$ (for some $m \in \mathbb Z)$

$\Rightarrow (b-4)^2 -16=m^2$

$\Rightarrow (b-4+m)(b-4-m)=16$

Therefore the possible cases are $b-4+m=\pm 8$, $b-4-m=\pm 2$,$b-4+m=b-4-m=\pm 4$

i.e b-4=5,-5,4,-4

$\Rightarrow b =9,-1,8,0$

Therefore $ (b^2)_{max} = 81$

Arrangement Problem | AIME I, 2012 | Question 3

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

Arrangement - AIME 2012

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

is 107
is 216
is 840
cannot be determined from the given information

Key Concepts

Arrangements

Algebra

Number Theory

Check the Answer

Answer: is 216.

AIME, 2012, Question 3

Combinatorics by Brualdi

Try with Hints

Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions

one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.

Then $(3.2)^3={216}$ways

Problem related to Money | AMC 8, 2002 | Problem 25

Try this beautiful problem from AMC-8, 2002 related to money (problem 25).

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?

$\frac{1}{3}$
$\frac{1}{4}$
$\frac{3}{4}$

Key Concepts

Algebra

Number theory

fraction

Check the Answer

Answer:$\frac{1}{4}$

AMC-8 (2002) Problem 25

Challenges and Thrills in Pre College Mathematics

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Each Friend gave Ott the equal amount of money

Can you now finish the problem ..........

Assume that ott gets y dollars from each friend

Can you finish the problem........

Given that Ott gets equal amounts of money from each friend,
we can say that he gets y dollars from each friend.
This means that Moe has 5y dollars,
Loki has 4y dollars, and Nick has 3y dollars.
The total amount is 12y dollars,
Therefore Ott gets 3y dollars total,
Required fraction =$\frac{3y}{12y} = \frac{1}{4}$

Divisibility Problem | PRMO 2019 | Question 8

Try this beautiful problem from the PRMO, 2010 based on Divisibility.

Divisibility Problem - PRMO 2019

Find the number of positive integers such that $3 \leq n \leq 98$ and $x^{2^{n}}+x+1$ is divisible by $ x^{2}+x+1.$

is 107
is 48
is 840
cannot be determined from the given information

Key Concepts

Inequalities

Algebra

Number Theory

Check the Answer

Answer: is 48.

PRMO, 2019, Question 8

Elementary Number Theory by David Burton

Try with Hints

for n={3,5,...97}

where n is odd since factor of $x^{2}+x+1$ is also factor of given expression

Then n=48.

Theory of Equations | AIME I, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

Theory of Equations - AIME 2015

Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

is 107
is 72
is 840
cannot be determined from the given information

Key Concepts

Series

Theory of Equations

Algebra

Check the Answer

Answer: is 72.

AIME, 2015, Question 10.

Polynomials by Barbeau.

Try with Hints

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

Trigonometry Problem | AIME I, 2015 | Question 13

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

Trigonometry Problem - AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n\), where (m) and (n) are integers greater than 1. Find (m+n).

is 107
is 91
is 840
cannot be determined from the given information

Key Concepts

Trigonometry

Sequence

Algebra

Check the Answer

Answer: is 91.

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 - 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.

Smallest Perimeter of Triangle | AIME I, 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

Smallest Perimeter of Triangle - AIME 2015

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$..

is 107
is 108
is 840
cannot be determined from the given information

Key Concepts

Inequalities

Trigonometry

Geometry

Check the Answer

Answer: is 108.

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

Try with Hints

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.Now let $BD=y$, $AB=x$, and $\angle IBD$ =$ \frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \frac{y^2-32}{32}$.

Cross-multiplying yields $32y = x(y^2-32)$.

Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$.

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = {108}$.

Cube of Positive Integer | Number Theory | AIME I, 2015 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

Cube of Positive Numbers - AIME I, 2015

There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

is 107
is 183
is 840
cannot be determined from the given information

Key Concepts

Algebra

Theory of Equations

Number Theory

Check the Answer

Answer: is 183.

AIME, 2015, Question 3

Elementary Number Theory by David Burton

Try with Hints

$a^{3}=12p+1$ implies that $a^{3}-1=12p$ that is (a-1)($a^{2}$+a+1)=12p

a is odd, a-1 even, \(a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13

\(a^{2}+a+1=p implies p= 169+13+1=183.

Algebra based on Integer PRMO Problem 6

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Mixture | AMC-8, 2002 | Problem 24

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Algebra based on Quadratic equation PRMO Problem 9

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Arrangement - AIME 2012

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Problem related to Money - AMC-8, 2002 - Problem 25

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Divisibility Problem - PRMO 2019

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Theory of Equations - AIME 2015

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Trigonometry Problem - AIME 2015

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Smallest Perimeter of Triangle - AIME 2015

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Cube of Positive Numbers - AIME I, 2015

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