TheXSquaredFactor

Neat! Very clever, indeed.

Let's think about this problem; I've been begged to attempt to solve this problem, so I guess I will give it a shot:

\(\frac{x}{4000-x}=a^2\)

This represents the problem exactly.

Let's solve for x:

Ok, now let's try and think about this problem logically. \(a^2\) will never be divisible by \(1+a^2\) because adding one to a number means that the numbers are co-prime. Because of this, we do not have to consider any of the cases. The only time there will be integer solutions for x is when 4000 is divisible by \(1+a^2\); in other words, we need to know the factors of 4000. 

The factors of 4000, in order, are \({1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,500,800,1000,2000,4000}\). Now, we must set the denominator to all these values and solve:

First is 1. Here we go:

Here's a question, though. Is 0 a perfect square? Unfortunately, I am unsure about whether or not it is indeed a perfect square because definitions vary. Many people agree with the definition of if \(\sqrt{a}\in\mathbb{Z}\)where \({a}\in\mathbb{Z}\) , then that number is a perfect square. With that definition, 0 is a perfect square. However, some define it as \(\sqrt{a}\in\mathbb{N}\) where \({a}\in\mathbb{Z}\). In this case, 0 would be excluded. It's an open question, really, without a concrete answer. If you know whether or not an answer exists to this question, I would appreciate feedback. Anyway, let's try the next factor, 2:

You know what? There is a method to make this go even faster, anyway! Let's take the following set and subtract 1 from every term.

Now,\({1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,500,800,1000,2000,4000}\)

transforms into \({0,1,3,4,7,9,15,19,24,31,39,49,79,99,124,159,199,249,399,499,799,999,1999,3999}\). Now, out of those, which ones are perfects squares? They are the following:

\(0,1,4,9,49\)

Now, take the square roots of the numbers in the set:

\(0,1,2,3,7\)

Therefore, the sum of the positive perfect square integers is \(0+1+2+3+7=13\).

The question does not ask for the solution of the system of equations but rather how many solutions exist. There are only 3 possibilities. I will list them for you.

1. No solution

2. One solution

3. Infinitely many solutions

Currently, both equations are written in the form of \(Ax+By=C\). However, in a system, the system looks like the following:
 

{ \(A_1x+B_1y=C_1\)

{\(A_2x+B_2y=C_2\)

Now, let's compare the ratios of the system

{\(2x+3y=20\)

{\(6x-y=20\)

When I say compare the ratio, I mean that you want to look at the relationship of \(\frac{A_1}{A_2}\) and \(\frac{B_1}{B_2}\) and \(\frac{C_1}{C_2}\).

Let's look at them:

\(\frac{A_1}{A_2}=\frac{2}{6}=\frac{1}{3}\)

\(\frac{B_1}{B_2}=\frac{3}{-1}=-3\)

\(\frac{C_1}{C_2}=\frac{20}{20}=1\)

Since \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), there is only one solution.

This is a rare example in mathematics where, I believe, parentheses is necessary in order to evaluate the expression without ambiguity. I will demonstrate why.

Strictly speaking, asinus's interpretation is incorrect. If you were to evaluate this with a calculator inputted like as is, the calculator would evaluate it as \(\frac{6^2}{2}*3+4\). This is because the 2(3) is really multiplication, so division takes precedence since it is comes first in the expression. First it does 6^2, then it divides 6^2 by 2 because division is first from left to right, and then it multiplies that quantity by 3. Here is another example with a variable 

8/2y

Using the same logic as above, this equation, in fraction form is strictly \(\frac{8}{2}y\)--not \(\frac{8}{2y}\). Some would argue, however, that 2y is a term, so it shouldn't be separated.  

How do we eliminate this ambiguity if there is no fraction button to speak of? Use parentheses! 

Asinus's interpretation of \(\frac{6^2}{2*3}+4\) will be unambiguous once you add 1 set of parentheses with \(6^2/(2(3))+4\). Now, the only correct interpretation is \(\frac{6^2}{2*3}+4\) because the parentheses indicate that we are dividing by the quantity of the product of 2 and 3.

The strict interpretation is \(\frac{6^2}{2}*3+4\) should be written like \((6^2/2)(3)+4\). In this case, the quantity of six squared divided by two is all multiplied by three. No more ambiguity.

Okay, after all of this ranting, now I will evaluate what I believe to be, under the current rules of the order of operations, the way to evaluate the expression 6^2/2(3)+4 as \(\frac{6^2}{2}*3+4\):

There are many ways to prove that a triangle is right, but this method, I believe, is the best and most efficient. Our first task is to find the actual length of the sides. To do this, I will utilize a fomula called the distance formula. The formula is the following:

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Let's use this to find the distances:

Do you remember that famous formula called the Pythagorean theorem? It describes the relationship of the sides of a right triangle, stating that the sum of the squares of the legs is equal to the hypotenuse squared:

\(a^2+b^2=c^2\)

The converse happens to also be true; in other words, if \(a^2+b^2=c^2\), then the triangle is right. Let's check to see if this condition is true.

Before we do, however, we need to understand which sides are the legs and which is the hypotenuse. The hypotenuse is always the longest side of the triangle. Although we do not know the exact length of the segments AB, BC, and CA, we can determine which side is the longest since we know that \(AB=\sqrt{53}\), \(BC=\sqrt{106}\), and \(CA=\sqrt{53}\). Logically speaking,  \(\sqrt{106}>\sqrt{53}\) because the higher the radicand, the larger the actual value is. Now that we understand the hypotenuse is \(BC\), the legs do not matter. One is assigned the value a and one is assigned the value b. To confirm that this triangle is indeed right, plug in the values into the equation:
 

Because \(a^2+b^2=c^2\), we can conclude that this triangle is indeed right.

A convex, 11-sides polygon (known as a hendecagon) may only have, at most, 3 acute interior angles.

This is because each acute angle results in an exterior angle that is greater than 90 degrees. If you were to add a fourth exterior angle, the sum of all the exterior angles will be greater than 360 degrees. Of course, the sum of all the exterior angles in a convex polygon must be equal to 360, so if the sum of all are greater than this number, you can cannot create a convex polygon. 

Wow! That is such a mouthful of an expression, but I will attempt to evaluate it anyway. 

If I am not mistaken, the expression is the following:
 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\)

At this point, I cannot simplify the exponent any further. I have done an impressive job of simplifying 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\) into \(61^{({62^{30}*\sqrt{3905}})}\). I will now have to utilize an extremely accurate calculator for a calculation of this size. My only option is to provide you with an exponent tower; that's how huge it is!

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}=61^{({62^{30}\sqrt{3905}})}\approx 10^{10^{55.81927966668297}}\)

Questions like this one appear to be asked frequently on this forum, so I will attempt to break it down for you so that you are not confused at all.

1. Identity Place Value The Decimal Extends To

\(0.3456\) extends until the ten thousandths. That is all this step entails. 

2. Take the Decimal Value over That

Since our decimal extends 4 decimal places, we want to make it equivalent to that number over ten thousand.

\(0.3456=\frac{3456}{10000}\)

3. Find the GCF of the Numerator and Denominator and Divide by It

GCF stands for "greatest common factor." Your goal in this step is to find what that is. If it is difficult to determine it, like it is here, you may want to try to reduce the fraction little by little.

For example, 3456 and 10000 are both even, so it is divisible by 2

However, someone more observant might notice that the last two digits of both numbers, 56 and 00, are divisible by 4, so the GCF is, at least, 4.

\(\frac{3456}{10000}\div\frac{4}{4}=\frac{864}{2500}\)

Yet again, 64 and 00 are divisible by 4, so divide it by 4 again.

\(\frac{864}{2500}\div\frac{4}{4}=\frac{216}{625}\)

4. You're done!

You will have a field day with this