Display Problems

Imagination is more important than knowledge... Albert Einstein

Guess is more important than calculation --- Knowhowacademy.com

Polynomials - Factorization and modification

For COMPETITION
Number of Total Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10

Type:

Section:Polynomials

Theme:
Adjustment# : 0

Difficulty: 1

'
Let $a$ , $b$ , and $c$ be positive integers with $age$ $bge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$ .
What is $a$ ?
$extbf{(A)} 249qquad extbf{(B)} 250qquad extbf{(C)} 251qquad extbf{(D)} 252qquad extbf{(E)} 253$
'

Category Factorization and modification

Analysis

Solution/Answer
Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$ .
Now, this can be rearranged:
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
and factored:
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ , $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that $14 = 9 + 4 + 1$ .
$(a-c)^2 = 9 ightarrow a-c = 3$ , since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 2$ or $1$ , or whether $b-c = 1$ or $2$ .
We want to solve for $a$ , so take the two cases and solve them each for an expression in terms of $a$ . Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$ . Plug these values into one of the original equations to see if we can get an integer for $a$ .
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$ , after some algebra, simplifies to $7a = 2021$ . 2021 is not divisible by 7, so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$ , which simplifies to $8a = 2024$ . Thus, $a = 253$ and the answer is $oxed{ extbf{(E)} 253}$ .

Answer:

Problem Num : 2

From : AMC10B

Type:

Section:Polynomials

Theme:
Adjustment# : 0

Difficulty: 1

'
Let a, b, and c be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$mathrm{(A) }0qquadmathrm{(B) }1qquadmathrm{(C) }4qquadmathrm{(D) }7qquadmathrm{(E) }8$
'

Category Factorization and modification

Analysis

Solution/Answer
$a+8c=7b+4$ and $8a-c=7-4b$
Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.
Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=1$ . Answer choice $oxed{(B)}$ .

Answer:

Problem Num : 3

From : NCTM

Type: Complex

Section:Polynomials

Theme:Manipulation
Adjustment# : 0

Difficulty: 3

Category Factorization and modification

Analysis

Solution/Answer

Problem Num : 4

From : AMC10B

Type:

Section:Polynomials

Theme:
Adjustment# : 0

Difficulty: 1

'
Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that

$P(1) = P(3) = P(5) = P(7) = a$ , and
$P(2) = P(4) = P(6) = P(8) = -a$ .

What is the smallest possible value of $a$ ?
$extbf{(A)} 105 qquad extbf{(B)} 315 qquad extbf{(C)} 945 qquad extbf{(D)} 7! qquad extbf{(E)} 8!$
'

Category Factorization and modification

Analysis

Solution/Answer
There must be some polynomial $Q(x)$ such that $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$
Then, plugging in values of $2,4,6,8,$ we get
$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a$ $P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a$ $P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a$ $P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a$
$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $lcm(15,9,15,105)$ . Solving, we receive $315$ , so our answer is $oxed{ extbf{(B)} 315}$ . $lacksquare$

Answer:

Array ( [0] => 7940 [1] => 7980 [2] => 3011 [3] => 8140 ) 4