All Answers 354538 Answers

n mod 7 = 1, n mod 8 = 4, n mod 9 = 7, solve for n

Using "Chinese Remainder Theorem + Modular Multiplicative Inverse", we get:

n = 504m + 484, where m = 0, 1, 2, 3.......etc. 

I'm part of an AoPS class, and I've seen other people spam AoPS questions.

I've never done this personally, but I've had people literally joking in my class about web2.0calc.

One person would say, "You know there's like an answer key to AoPS questions online?"

Another would say, "Yeah! web2.0calc is epic!"

And they just start laughing.

Divisibility rules of 8 and 9 are easiest, so let's find that first.

Modulo may be a bit hard for now, so I'll show you a more rudimentary way.

The list of numbers that when divided by 9 give a remainder of 7 are 7, 16, 25, 34, 43, 52

Same for 8 that give 4: 4, 12, 20, 28, 36, 44, 52

We see that 52 meets both of these requirements! This will happen every cycle of 72, because the LCM of 8 and 9 is 72.

(in other words, 52, 124, 196, 268, 340, ...)

Now, the three digit numbers are 124, 196, 268, etc.

We still haven't used the divisible by 7 with remainder of 1 rule!

Those numbers are 1, 8, 15, 22, etc.

If we test every single number that we had before (124, 196, 268, 340, 412, 484, 556, 628, 772, 854, 916, 988), we can find that

484 can be divided by 7 to give a remainder of 1. So can 988.

Therefore, there are two numbers - 484 and 988.

There is a faster way to do this - with modulo, but try looking at that for yourself! Search up "modulo" online and start looking at the math sites.

You are very welcome!

:P

(x+8)(x^2+6)

There isn't really a clear way of solving cubic equations. You can use any math calculator online, but if you're on paper, test some values. You can find factors of 48, then try from there.

You are very welcome!

:P

I assume you mean \(\sqrt{p+a+r}=k\).

Whenever there is a square root symbol, you want to square the equation. 

\((\sqrt{p+a+r})^2=k^2\)

This becomes \(p+a+r=k^2\), because the square root symbol is lifted.

Now we can add and subtract values on each side of the equation to find a. I think you can do that.

You are very welcome!

:P

straight line depreciation.....   the same $ amount per year....  NOT a prcentage r exponential amount

CalculatorUser, regarding your "Wonderman observed in this class that some people were cheating their homework, and not actually learning."   

Those students, who are not learning, will not be able to come here and ask for the solutions when they're taking their exams.  If they've routinely depended on this site to do their homework for them, and not learned from the examples which were solved, then the abrupt drop in the quality of their work on the exams will betray that they've been cheating.  It will also drop their grade point.  The habitual cheaters will be winnowed out by a sort of natural selection.

I solve a problem here occasionally.  I'm not good enough to do the higher maths like calculus and abstract algebra, but I can do relatively simpler problems.  I hope my meager contributions help free up the time of the really smart responders, like CPhill and heureka, and the omission of a mention of other superstars is not meant as a slight, I just picked the first two I thought of. 

The only way I know to help is to solve the problem.  Hopefully, seeing the solution will instill learning by example.  I'm not a teacher, I don't have teacher training, and I wouldn't know how to start teaching.  I remember when I was in school, if I was stuck and somebody solved a problem for me, most of the time that would turn on the light above my head.  By seeing one problem solved, I learned to solve that type of problem.

I'm certain that wonderman's intentions are good, it's just that continually griping at me isn't going to get him anywhere.  The responders here have good intentions, too.  Instead of criticizing me/others for our good intentions, an alternative for him/her could be to lead by example.  Even if nobody follows, that would create one new leader.

.

lolololol, but they use bad words

Why is it a 'challange questions'?

\(\cos(2\theta) = 2\cos(\theta)\\ cos^2\theta-\sin^2\theta = 2\cos(\theta)\\ cos^2\theta-(1-\cos^2\theta) = 2\cos(\theta)\\ 2cos^2\theta-1 = 2\cos\theta\\ 2cos^2\theta-2\cos\theta-1 = 0\\\)

Now let x=cos (theta)  and solve for x.

etc

There is a reason as to why no one has answered your question, me included. First, you have posted this problem twice. Second, the use of words in your title, HELP PLZ HELP suggest some very urgent help. Why would you need urgent help if you're not cheating. Because of this, most Web2.0calc users will assume that you are cheating. Lastly, you have, frankly, quite a rude message at the bottom of your question, " I need help!!!!! ". Notice the use of exclamation marks. To be honest, there's no reason for us to answer your question.

YES I will second that. 

Thanks Tom...riddle

Hi MagicKitten,

I saw you ask how this forum should be used. (earlier today)

How you and and Tom worked together on this thread is EXACTLY how it should be used.

To bad the asker was not involved but it doesn't matter.  It was fabulous that you stepped in. :)

I believe that there are some things that Wonderman can do instead of downvoting the question. He could, give help and not the answer or  ask people to give help instead of giving the answer. I think this way, he will make a much more positive impact on the community instead of being a police officer. However, there is nothing we can do about the cheaters, in my opinion. If they cheat, it's their fault. We don't lose anything from answering or giving help to the question. They lose because they fail to learn the process and how to do the problem. We can't let cheating negatively affect this forum.

Thank u sm <3 

Thank you, CPhill !

Find the sum of the numbers \(3,8,13,\ldots,253\)

arithmetic series:

\(\mathbf{n=\ ?}\)

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)\cdot d \quad &| \quad a_n = 253,\ a_1=3,\ d=5 \\\\ 253 &=& 3+(n-1)\cdot 5 \quad &| \quad -3 \\ 250 &=& (n-1)\cdot 5 \quad &| \quad :5 \\ 50 &=&n-1 \quad &| \quad +1 \\ 51 &=&n\\ \mathbf{ n } &=& \mathbf{51} \\ \hline \end{array}\)

sum:

\(\begin{array}{|rcll|} \hline sum &=& \left(\dfrac{a_1+a_n}{2} \right) \cdot n \quad &| \quad a_n = 253,\ a_1=3,\ n=51 \\\\ sum &=& \left(\dfrac{3+253}{2} \right) \cdot 51 \\ sum &=& \left(\dfrac{256}{2} \right) \cdot 51 \\ sum &=& 128 \cdot 51 \\ \mathbf{sum} &=& \mathbf{6528} \\ \hline \end{array}\)

This is a partitions problem in 4 parts:
9 + 9 + 9 + 3 = 30...........4 permutations
9 + 9 + 8 + 4 = 30...........12  
9 + 9 + 7 + 5 = 30...........12
9 + 9 + 6 + 6 = 30...........6
9 + 8 + 8 + 5 = 30...........12
9 + 8 + 7 + 6 = 30...........24
9 + 7 + 7 + 7 = 30............4
8 + 8 + 8 + 6 = 30...........4
8 + 8 + 7 + 7 = 30...........6
If you add up all the permuations, you should get =84

Not only that, strange things have happened due to this cheating. 

Thanks Max,

It is good to see you again :)

Actually I was scrolling through some old posts, and Melody was the first one to see that.

I just used Melody's observation really.

wonderman, I looked at your answers history, and this is the only thing you ever post. 

I think you could be a great help to some of these people, instead of just being a scold.

calculator user, that was an excellent observation. 

.

I'm going to explain ElectricPavolv's answer, which is indeed correct. He got the answer using his knowledge of difference of squares, which states that a quadratic equation in the form of a^2-b^2 = (a+b)(a-b). In this case, we substitute k for a and 9 for b, to get a factorization fo (k-9)(k+9).

sort of like a self-instated soldier or fighter, or someone that stops crime.

Someone like batman, no one ever told him to be a superhero, but he just became a superhero and started fighting the drug dealers like an unofficial cop.

YES TOM MARVOLO RIDDLE!

This is how problems should be solved on this site!

Welcome! Here at the bottom right you can see a topic that says Sticky Topics. Click on the top which says Welcome To Web2.0calc where there is a lovely welcome message and rules and guidelines. Have fun!