AMC 10A 2021 Problem 9 | Factorizing Problem

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Factorizing Problem: AMC 10A 2021 Problem 9

What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$ ?

0
$\frac{1}{2}$
$\frac{1}{4}$
1
2

Key Concepts

Expansion of Polynomial

Factorization

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AMC 10A 2021 Problem 9

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Expand the expression.

So, we get that the expression is $x^{2}+2 x y+y^{2}+x^{2} y^{2}-2 x y+1$ or $x^{2}+y^{2}+x^{2} y^{2}+1$.

Then the minimum value for this is 1 , which can be achieved at $x=y=0$. .

Pairs of Positive Integer | ISI-B.stat | Objective Problem 178

Try this beautiful problem based on Pairs of Positive Integer, useful for ISI B.Stat Entrance.

Pairs of Positive Integer | ISI B.Stat Entrance | Problem 178

How many pairs of positive integers (m, n) are there satisfying $m^3 – n^3 = 21$?

(a) Exactly one
(b) None
(c) Exactly three
(d) Infinitely many

Key Concepts

Integer

Factorization

Odd number

Check the Answer

Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

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Given that $m^3– n^3 = 21$

$\Rightarrow (m – n)(m^2 + mn + n^2) = 3*7$

Possible cases are , $m – n = 3$, $m^2 + mn + n^2 = 7$ and $m – n = 1$, $m^2 + mn + n^2 = 21$

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

Now according to the questions we are trying to find out the positive integers,so we will neglect the negetive cases..........

First case, $(3 + n)^2 + (3 + n)n + n^2 = 7$
$3 n^2 + 9n + 2 = 0$
$n = \frac {-9 \pm \sqrt {9^2 – 432}}{6}$= not integer solution.
So this case is not possible.
Second case, $(n + 1)^2 + (n + 1)n + n^2 = 21$
$\Rightarrow 3n^2 + 3n – 20 = 0$
$n = \frac {-3 \pm \sqrt {9 + 240}}{6}$ = not integer solution.

Therefore option (b) is the correct

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

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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 $a^2 + b^2 + c^2=17+1=18$

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

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(‘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$

AMC 10A 2021 Problem 9 | Factorizing Problem

Factorizing Problem: AMC 10A 2021 Problem 9

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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Pairs of Positive Integer | ISI-B.stat | Objective Problem 178

Pairs of Positive Integer | ISI B.Stat Entrance | Problem 178

Key Concepts

Check the Answer

Try with Hints

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Problem based on Integer | PRMO-2018 | Problem 6

Algebra based on Integer PRMO Problem 6

Key Concepts

Check the Answer

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Quadratic Equation Problem | PRMO-2018 | Problem 9

Algebra based on Quadratic equation PRMO Problem 9

Key Concepts

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Factorizing Problem: AMC 10A 2021 Problem 9

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Pairs of Positive Integer | ISI B.Stat Entrance | Problem 178

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Algebra based on Integer PRMO Problem 6

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Algebra based on Quadratic equation PRMO Problem 9

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