AMC 8 2019 Problem 20 | Fundamental Theorem of Algebra

Try out this beautiful algebra problem number 2 from AMC 8 2019 based on the Fundamental Theorem of Algebra.

AMC 8 2019 Problem 20:

How many different real numbers $x$ satisfy the equation $(x^{2}-5)^{2}=16?$

$\textbf{(A) }0$
$\textbf{(B) }1$
$\textbf{(C) }2$
$\textbf{(D) }4$
$\textbf{(E) }8$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is (D) 4

AMC 8, 2019, Problem 20

Try with Hints

The given equation is

$(x^2-5)^2 = 16$

and that means

$x^2-5 = \pm 4$

Among both cases, if

$x^2-5 = 4$

then,

$x^2 = 9 \implies x = \pm 3$

and that means we have 2 different real numbers that satisfy the equation.

and if we take another case, then

$x^2-5 = -4$

and so,

$x^2 = 1 \implies x = \pm 1$

and that means we have 2 different real numbers in this `case too that satisfy the equation. So total 2+2=4 real numbers that satisfy the equation.

AMC 8 2019 Problem 16 | Algebra Problem

Try this beautiful Number Theory problem from the AMC 2019 Problem 16. You may use sequential hints to solve the problem.

Algebra Question - AMC 8, 2019 Problem 16

Qiang drives 15 miles at an average speed of 30 miles per hour. How many additional miles will he have to drive at 55 miles per hour to average 50 miles per hour for the entire trip?
(A) 45
(B) 62
(C) 90
(D) 110
(E) 135

Key Concepts

Algebra

Value

Average Speed

Check the Answer

Answer: is (D) 110

AMC 8, 2019, Problem 16

Try with Hints

Among the options, there is only one option which is divisible by 55 and that is 110.

That option tells the travel hour is 2.

Qiang drives 15 miles at an average speed of 30 miles per hour.

And we know, Average speed = Total Distance/Total Time

So, by the formula,

$\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$

In this case, If we consider the whole journey, Total Distance is (110+15)=125

And as Qiang has to drive at 50 miles per hour for the entire trip, and as Average speed = Total Distance/Total Time ,

$\frac{125}{50} = \frac{5}{2}$

As both are same , our answer 110 is established.

AMC 8 2019 Problem 17 | Value of Product

Try out this beautiful algebra problem from AMC 8, 2019 based on finding the value of the product. You may use sequential hints to solve the problem.

AMC 8 2019: Problem 17

What is the value of the product

$\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right) ?$

(A) $\frac{1}{2}$

(B) $\frac{50}{99}$

(D) $\frac{100}{99}$

(E) $50$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is $\frac{50}{99}$

AMC 8, 2019, Problem 17

Try with Hints

We write

$\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$

in a different form like

$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$

All of the middle terms eliminate each other, and only the first and last term remains i.e.

$\frac{1}{2} \cdot \frac{100}{99}$

$\frac{1}{2} \cdot \frac{100}{99}=\frac{50}{99}$

and that is the final answer.

Perfect square and Positive Integer | TOMATO B.Stat Objective 115

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Perfect square and Positive Integer.

Perfect Square & Positive Integers ( B.Stat Objective Question )

If n is a positive integer such that 8n+1 is a perfect square, then

n must be odd
2n cannot be a perfect square
n must be a prime number
none of these

Key Concepts

Perfect square

Positive Integer

Primes

Check the Answer

Answer: 2n cannot be a perfect square.

B.Stat Objective Problem 115

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

Let $(8n+1)=k^{2}$ be a perfect square so k is found to be odd here as 8n+1 is odd.

$\Rightarrow 8n=k^{2}-1$

$\Rightarrow 8n=(k-1)(k+1)$

$\Rightarrow 2n=\frac{(k-1)(k+1)}{4}$

$\Rightarrow (\frac{k-1}{2})(\frac{k+1}{2})$

here (k-1) and (k+1) are consecutive even numbers then $(\frac{k-1}{2})(\frac{k+1}{2})$ are consecutive even numbers

$2k \times (2k+2)= 2^2{k(k+1)}$ is not a perfect square as product of two consecutive numbers proved here as not a perfect square

So, 2n is not perfect square.

Smallest Positive Integer | PRMO 2019 | Question 14

Try this beautiful problem from the PRMO, 2019 based on Smallest Positive Integer.

Smallest Positive Integer - PRMO 2019

Find the smallest positive integer n$\geq$10 such that n+6 is a prime and 9n+7 is a perfect square.

is 107
is 53
is 840
cannot be determined from the given information

Key Concepts

Integers

Primes

Perfect Square

Check the Answer

Answer: is 53.

PRMO, 2019, Question 14

Elementary Number Theory by David Burton

Try with Hints

Let 9n+7=$m^{2}$ n+6 prime then n+6 odd then n is odd then n=2k+1 then 9(2k+1)+7=$m^{2}$ then 18k=$m^{2}$-16=(m+4)(m-4) then 18k even m is even then m=2p

18k=(2p+4)(2p-4)=4(p+2)(p-2) then 9k=2(p+2)(p-2)then k even then k=2d then 18d=2(p+2)(p-2) then 9d=(p+2)(p-2) then p of form 9q+2,9q-2

for p=9q-2 then m=2(9q-2) for q=1 then$m^{2}$=196then n=21 then n+6=27 non prime, for p=9q+2 then m=2(9q+2) for q=1 $m^{2}$=484 then n=53 then n+6=59 prime then n=53.

AMC 8 2019 Problem 20 | Fundamental Theorem of Algebra