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ibhar:

Where is your response to Textot?     

Answerers want interaction!

Thanks Textot   

Alan proved what you asked and you did not even bother to thank him.

You did not even give him a point.

Why would any serious answerers bother with you?

Hi jsaddern and catmg,     

I will try and explain how I was thinking.

I know this.

\((10^{n+1}+1)>(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)

So I know that if    \((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)   is a perfect sqare then

\((10^{n+1}+1)\)     must have at least one perfect square factor.   Can you see that?

So let         \((10^{n+1}+1) =a^2b\)         where a and b are rational positve numbers

then   \((a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\)    must be     bc^2   where b and c are rational positive numbers

and  a>c

That would make    \((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)= a^2b*bc^2=(abc)^2\)

So I googled online for a number of the form  10^n+1  with a perfect square factor.

I found this site:

Yes, it does include \(0\) and \(9\).

Also, yes! It looks like \(11\) does disprove it. Can you describe the set of all numbers for which the given expression is a perfect square? (Maybe to get there, how did you come up with your example?)

Find the length UV.

Hello Guest!

The length of the diagram \(\overline{QR}\)  is a.

\(\overline{ST}=2\cdot \overline{TR}+5\\ \overline{ST}+\overline{TR}=30\\ 2\cdot \overline{TR}+5+\overline{TR}=30\\ \overline{TR}=\frac{25}{3}\)

\(f(x)= -\frac{30-\frac{25}{3}}{a}\cdot x+30 =-\frac{65}{3a}\cdot x+30\\ g(x)= \frac{30}{a}\cdot x \\ f(x)=g(x)\\-\frac{65}{3a}\cdot x+30 = \frac{90}{3a}\cdot x\\ \frac{155}{3a}\cdot x=30\\ x=\frac{90a}{155}=\frac{18a}{31}\)

\(g(\frac{18a}{31})=\overline{UV}=\frac{30}{a}\cdot\frac{18a}{31}=17.42\)

\( The\ length\ \overline{UV}\ is\ 17.42\)

  !

I was also quite confused at the case when n = 10. 

I believe your example fits all the requirements. 

I'm pretty sure between 0 - 9 includes 0 and 9. 

=^._.^=

I think I can disprove it with an example.   n=10    (Edit correction, I had previously said n=11)

10^11+1 = 121*826446281

so

\(\sqrt{(10^{11}+1)*(826446281)} = \sqrt{121^2*826446281^2}=121*826446281 \)

\((10^{n+1}+1)(a_n \cdot 10^{n} + a_{n-1} \cdot 10^{n-1} + \ldots + a_0 \cdot 10^0)\\ when\;\; n=10\\ (10^{11}+1)(826446281)\qquad \text{is a perfect square}\)

sqrt((10^11+1)*826446281 = 9090909091

-------------------------------

Again I reread your question, it says 'between 0 and 9'  Is that supposed to include 0 and 9?

My answer uses 0.

I suppose I wasn't meant to but is 9 also really not allowed? 

period = 3 months     periods in 10 years = 40

interest per period   .04 / 4

5000 = (deposit) * (1 + .04/4)⁴⁰             solve for 'deposit'

Periods ina year = 12

interest per period    .075/12

periods = 120

FV = 300( 1 + .075/12)¹²⁰   = $ ___________

Sum of internal angles =  (n-2)* 180 = 540

Subtract the three 90 degree angles     540 - 3 * 90 = 270

270 degrees is two remaining angles total        each   135 degrees

DFO627 (DumbFuckOnline627):

Melody is not a dude; she’s a teacher. I’m pretty sure this is not the response Melody was looking for.

You titled you question with “Help me please!!!” and you concluded it with, “Can someone write this out step by step to the answer? I'd like to try to understand...”

Melody spent a fair amount of time answering and explaining the solution to your question. She’d like to know if you understood her solution and explanation. This is reasonable. Most teachers like to know if their presentations and instructions are elevating their students above dumbfuck status.

In this case, it’s a classic example of trying to teach a pig to sing. Trying to teach a pig to sing is a waste of time and always annoys the pig.  As anyone may see the pig was very annoyed and presented as a rude little asshole: “Dude, I don't have to reply to every post!”

It wasn’t a complete waste of time though; others will read this and learn from it.  The readers will learn some math ...and how not to be a rude little asshole.

It’s true; you don't have to reply to every post. It’s also true that you don’t have to be a rude little asshole. But you are what you are: a dumbfuck online and a rude little asshole. You chose that path... you are still young enough to change your path....

GA

--. .-

Write as an equation: 

\({2 \over 5} ={ x + 1000 \over 2x+ 1000}\)

Solving for x, we find John originally had \(\color{brown}\boxed{-3,000}\) dollars. 

Here's another way to prove it's impossible:

\(k \times 1984 \times 2 = k \times 2^7 \times 31\)

Using this logic, you can create the factors: 1, 2, 4, 8, 16, 31, 32, 62, 64, 124, 128, 248, 496, 992, 1984, and 3968.

Because we want 21 factors, we need the number to be a perfect square. The smallest way to do this is by multiplying \(3968 \times 31 \times 2\). This yields 27 factors, way too big. Any bigger and you get more factors. 

Thus, this problem is impossible, like geno said (Note: If this problem was just \(k \times 1984\), we could multiply by 31, and get 21 factors. )

ສສສວງ

k·1984·2  can be written as  k·3968  =  k·27·31

27·31  has these 16 factors: 

        1, 2, 22, 23, 24, 25, 26, 27,

        31, 31·2, 31·22, 31·23, 31·24, 31·25, 31·26, and 31·27  

If k is a number such as 3, another 16 factors will be created (3 times each of the previous 16 factors).

If k is 2, two more factors are created, 28  and  31·28, for a total of 18 factors.

If k is 4 = 22, another two factors are created:  29  and  31·29, for a total of 20.

If k is 8 = 23, another two factors are created:  210  and  31·210, for a total of 22.

If you don't exclude 1 as a factor, I don't know how to get exactly 21.

4*2+x2=16

8 + x² = 16

x² = 8

x = ± √8