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HELP! ME!

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Find the sum of all the perfect squares (positive integers) of the form $$\frac{x}{4000-x}$$
where x is an integer

tertre  Aug 2, 2017
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#1
+1493
+1

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:

 $$\frac{x}{4000-x}=a^2$$ Multiply by 4000-x on both sides of the equation. $$x=a^2(4000-x)$$ Distribute the a^2 to both terms inside of the parentheses. $$x=4000a^2-a^2x$$ Add a^2*x to both sides of the equation. $$x+a^2x=4000a^2$$ Let's factor out an x from both of the terms on the left hand side of the equation. $$x(1+a^2)=4000a^2$$ Divide by 1+a^2 on both sides of the equation. $$x=\frac{4000a^2}{1+a^2}$$

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 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:

 $$1+a^2=1$$ Subtract 1 on both sides of the equation $$a^2=0$$ Take the square root of both sides $$|a|=0$$ Split your answers into plus or minus. $$a=0$$

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 $$0+1+2+3+7=13$$.

TheXSquaredFactor  Aug 3, 2017
#2
+79894
0

Nice, X2.....!!!!

CPhill  Aug 3, 2017

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