Need Math Help?

[Gates]

Math is such a wonderful subject. I wish this thread were more active. So, in the interest of some activity, I'm going to post the following fun-ish question.

There is a five-digit number A. With a 1 after it, it is three times as large as with a 1 before it. What is the number?

If you post an answer, include a collapse tag to avoid spoiling it.

I think I got it, it was pretty easy.

Spoiler

Let A be an integer such that 9999 < A < 100000.
A*10+1=3*(100000+A)
10A+1=300000+3A
7A=299999
A=42857

5 Digits? YES
428571=3*142857? YES

So A=48257
QED

 

[Kal]

Just for the sake of pedantry, "QED" does not really apply here, and your opening sentence is a little off. Not sure how much you care about proof-reading proofs (tee hee), but just thought I'd provide that bit of input.

Just for the sake of clarity, let's say "A with a 1 before it" means 1 is the leading digit, and "A with a 1 after it" means 1 is the right-most digit. In that case, your answer is correct. Yeah, it's a pretty easy problem. I'll pull out an olympiad book and see if I can find a more exciting problem later.

 

[Hot_ArmS]

I like that problem

moar pl0x

 

[Gates]

Yeah I know my proofs are sloppy, but I've finished all my requisite undergraduate proof classes and my career path doesn't require me to write any sophisticated proofs in the future, so I don't really care.

Spoiler

I thought it was important to clarify in the setup that A was an integer because there probably exists some decimal that fulfills these requirements - i.e. that 1.#####=3*.#####1. It's unlikely but possible. Not that I'm going to go out of my way to find it out...

I can totally understand being pedantic though because I'm really pedantic myself.

 

[#HBC | J]

Oh my goodness! There is a math help thread?! I am so happy I found this.

Okay, so I seriously need some help. I am trying to sign up for classes for college but I can't get into the math class I want due to me being 21 points on the placement test too low for it. I got a 64 and I need at least an 85 on it. The subject that stopped me from it is Elementary Algebra and I was a bit embarrassed at the score. I mean I have completely forgotten how to do square roots and pie and just I am trying to re-teach myself but it's really hard and I have the make-up test coming soon and it's my last chance to get into Math for Liberal Arts.

I just need help with Algebra which I haven't taken in about 2-3 years due to me accelerating in math/taking Pre-calc my Junior year. It's been since my 8th grade year since I took a regular algebra class and I'm a high school graduate now.

Any help would be appreciative! In fact, if someone could talk to me sometime on something to possibly help knock this into my head more I would be overly happy with that.

 

[Kal]

Gates, with regards to the setup, I mean more specifically where you only let A be a five-digit integer. Technically, this means that A is an arbitrary integer between 9999 and 100000. So instead you should write something like:

Let A be an integer such that 9999 < A < 100000 and that, with a 1 after it, it is three times as large as with a 1 before it.

In other words, you need to be careful to state all of your relevant requirements when you define A. In this case, no real confusion arises, so I'm just stating things for the sake of pedantry.

J, first, keep in mind that it's spelled "pi." Anyway, I will happily discuss any topic if you bring it up here. The best way to get this stuff back into your head is to solve relevant problems. Here is a website devoted to this book. While technically copyrighted, the copyright holder has made it freely available for academic use. I would suggest reading it thoroughly, answering as many of the problems in it as possible, then coming back for help. I won't give you any answers, but I'll do my best to walk you through so that you have a firm understanding of the material. The odd-numbered questions have their solutions available on the website anyway.

 

[#HBC | J]

Thank you so very much! You do not know how much I appreciate this!

Okay I will get to work on this website and start trying to do the problems by myself. If I run into any hiccups where I honestly don't know I will definitely come here. I just have a bad feeling it will probably be the first problem or so, but I will try to pull through it myself!

Edit: Oops, sorry about the pi thing. Haha that's slightly embarrassing.

 

[Kal]

It's ok. Good luck. Don't feel bad if you have to ask questions. I'm only recommending that you try them first yourself so that you can reinforce how to solve them. I have no problem just giving you the answers, I just don't think that would be particularly beneficial given your situation.

 

[PsychoIncarnate]

I haven't taken a math class in 7 years. Recently I switched degrees and this new one requires calculus (or pre-calculus...I forgot which)

I have to work my way back up there...I didn't remember a lot of things on the placement tests.

One thing I forgot was adding fractions with Square roots in them

it was something like

√75 + 5√2
___ -- ___
2 ------- 4

(Pretend it looks like a fraction, and remove the --, the forums removes spaces)

 

[Gates]

It's ok Psycho, I think understand what you're trying to say. I'm assuming it would be:

(SQRT(75)/2)+(5*SQRT(2)/4)

Do you remember the rules for fraction addition and square root addition? Because they both apply here. In order to add fractions, we need to ensure that both denominators are the same. We can do this by multiplying one of the fraction terms by a number equivalent to one in order to get the denominator equal to the other denominator. Once we have done that, we can rewrite the whole fraction as one fraction under a common denominator and add the terms in the numerator together. But since these terms are square root terms, there's a catch - You can only add numbers with a square root term if the number inside the square root is the same. Before we can do that though, we must factor out all of the perfect squares out of the number inside each square root term (look for numbers that are multiples of 4, 9, 16, 25, etc.). 2 is prime, so it's already in simplest form, but75 is divisible by 25, which is a perfect square. We can factor 25 out of 75 and take the square root of it to get 5*SQRT(3). However, this cannot be added to the other term since the numbers inside the square root are both in their simplest form and not the same. So, our final answer is:

(10*SQRT(3)+5*SQRT(2))/4

Mathematicians will have you leave your answer in square root form (i.e. Do not attempt to evaluate this in your calculator).

Here's a step-by-step look.

(SQRT(75)/2)+(5*SQRT(2)/4)    //Original Problem
(2*SQRT(75)/4)+(5*(SQRT(2)/4) //Same denominator
(2*SQRT(75)+5*SQRT(2))/4      //One denominator
(2*SQRT(25*3)+5*(SQRT(2))/4   //Factoring the square roots
(2*5*SQRT(3)+5*SQRT(2))/4     //The square root terms cannot be added together
(10*SQRT(3)+5*SQRT(2))/4      //Final Answer

And for reference, what did you change your major to exactly?

 

[PsychoIncarnate]

I figured the denominators would have to be the same, but I went about it all wrong. It's petty easy now that I see it. I just REALLY didn't remember you could do that to squares

 

[Elyssa Xey Hexen]

Speaking of square roots. Does anyone know of a way of calculating square roots, 3rd roots, and so on by hand?

 

[Gates]

Speaking of square roots. Does anyone know of a way of calculating square roots, 3rd roots, and so on by hand?

lol the only person who I know that ever learned that algorithm was a math professor of mine who's like 60+ years old.

I personally don't know it, but you might be interested in this.
http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation
Wikipedia can actually be surprisingly accurate when it comes to math information, as can Yahoo Answers.

 

[Kal]

Here are two rather famous questions in mathematics. If you're familiar with math history, then you probably have these answers memorized.

An integer n > 1 is prime whenever it has no divisors other than 1 and n. Note that 1 is not prime. Are there infinitely many primes? If yes, prove it. If not, how many primes are there?

A real number r is rational when it can be expressed as a ratio of integers. Let r = sqrt(2) be the real number whose square is 2. Is r rational? Prove your answer.

Here is a slightly more abstract problem. I'll post the necessary definitions to avoid assuming a background in abstract algebra.

A group is a set with a binary operation, *, on the group, satisfying three properties:

1) Associativity: for any a, b, c in the group, (a * b) * c = a * (b * c).
2) Identity: there exists an element e, called the identity, such that, for any a in the group, a * e = e * a = a.
3) Invertibility: for every element a in the group, there exists an element a^{-1}, called the inverse of a, such that a * (a^{-1}) = e.

For brevity, we will avoid writing *, and instead simply juxtapose elements. Thus, a * b is equivalent to writing ab.

Prove:

1) that the identity e is unique, i.e., that if e and e' are both identities, then e = e'.
2) that the inverse is unique, i.e., that if a^{-1} and r are both inverses of a, then a^{-1} = r.

If anyone gets the above (it's not terribly hard, but appreciating the definition and its subtlety can be difficult for a newby), I'll get into some more advanced group theory. I'll touch on some specifics of the group of integers (with addition as the operation; as an aside you can prove that the integer with multiplication as the operation is not a group!) and cyclic groups.

 

[Kole]

How can I find the range of the following function without graphing/using a calculator?:

f(x) = ((x-3)(x+2)^2)/((x-5)^3)

 

[Kal]

Hey Kole, sorry about the late response on this one. The first trick is to see that the function is unbounded; as x approaches 5, the numerator approaches 98, while the denominator approaches 0. Now, all you have to do is see what happens as the function approaches 5 from the left and from the right. From the left, we see that the function gets increasingly negative, as the numerator is positive, but the denominator is negative. From the right, it gets increasingly positive, as both the numerator and denominator are positive. If this isn't clear, just plug in a few values, like x = -4 and x = 6, for example.

Now, this function is clearly continuous (except at x = 5), which means that it will necessarily hit everything between any two y-values. To make this clear: if we draw this graph without lifting our pencil, and we start at -5 and end at +5, it's obvious that we will have to hit the values in between. Because there is a discontinuity at x = 5, we need to look at the range in two ways. From (-infinity, 5) and from (5, infinity), because we clearly must lift our pencil at x = 5.

We'll start with (-infinity, 5). Not sure on your calculus background, but it's not too hard to see that, as x gets increasingly negative, f(x) approaches 1. I can elaborate on this later if you do not see why. This means that (-infinity, 1) is in the range; if you start all the way at left-end of the x-axis, you'll be at 1, and you must draw all the way to -infinity without lifting your pencil. So, hopefully obviously, this means (-infinity, 1) is in your range.

A similar argument for (5, infinity) shows that (1, infinity) is in your range; you start at infinity, and draw all the way to 1.

Now you just need to show that f(x) is never 1, which is doable, albeit not easily. If you just care for a heuristic argument, note that the function is asymptotically approaching 1 from the left and the right; on the left, it is approaching below, and on the right, it is approaching above, and clearly never hits 1. This heuristic reasoning arguably requires a graph in the first place, so it might not be particularly useful. However, once you've shown that f(x) is never 1, it's clear that the range is just:

(-infinity, 1) U (1, infinity), i.e., the entire real line minus {1}.

Hope this helps. If anything is unclear or poorly reasoned, please let me know and I'll try to elaborate or fix the problem.

 

[Vinylic.]

I'm in need for a bit of help for remembering things:

Solving mathematic problems in 3 digits:
xxx * y

Multiplying or dividing fractions and those numbers like .1

Not even trolling here, every math problem I faced is just solved in high school was by calculator and steps from a math book owned by public schools. I would've done mathematics a little better I simply forgot how to multiply 3 or more digits by one digit without a calculator.

 

[Kal]

Let me know if this helps. If not, I'll write up something in TeX. But I think explaining it in the Smashboards forum would be difficult.

The trick in all of this is to just do integer multiplication and then move the decimal place later. For example, 3.77 x 2.8 is the same as (377 x 10^{-2})(28 x 10^{-1}), which is the same as 377 x 28 x 10^{-3}. This is why you just add the number of decimal place to get 3, and calculate the product 377 x 28 and move the decimal three places to the left. 377 x 28 is 10556, and 3.77 x 2.8 is 10.556, which is just 10556 with the decimal moved three places to the left.

 

[Elyssa Xey Hexen]

Out of curiosity, were you allowed to use calculators on exams? In all of my classes we were not allowed to use calculators for exams. This was something that continues to haunt me for the rest of my days in any mathematics class.

 

[Kal]

It varied from class to class for me. In high school, most of my classes allowed calculators.

 

[Chroma]

[COLLAPSE="Terse Solution:"]Let's assume the set's closed under *:
If e and e' are both identities, then e = e*e' = e' and
if a*a' = e = a*a'', (so a' and a'' are my two inverses, e's my identity), then
(a'')*(a*a') = (a''*a)*a' (associativity)
a''*e = e* a'
a'' = a'[/COLLAPSE]

I'm answering because I wanna encourage a talk about more group theory =)

 

[Kal]

Good answers, but they are a little terse. Nothing wrong with that, but with something this simple you really need to be careful. Also, thanks for pointing out that I forgot to mention that the operation must be closed. I thought I had written "binary operation," but did not.

In the future, we should write our answers in collapse tags.

 

[Chroma]

Good answers, but they are a little terse. Nothing wrong with that, but with something this simple you really need to be careful. Also, thanks for pointing out that I forgot to mention that the operation must be closed. I thought I had written "binary operation," but did not.

In the future, we should write our answers in collapse tags.

Thanks for teaching me about collapse tags! They're awesome-looking.

 

[Elyssa Xey Hexen]

Okay, lets declare some sort of closed system called *. Then we have two things which act as identities which we will name e and e'.

I guess I should assume e and e' belong to *? We were not exactly told this. If that is the case and since * is closed, then e*e' will exist within * by definition of it being a closed system. Am I suppose to be proving some result here? I do not really see a question being asked at all.

I am a bit confused about a' and a'' being inverses. Just from working backwards. You say a' and a'' are inverses. With this statement, a*a' = e = a*a'', the thing a is multiplied by either of these two inverses which brings it back to an identity statement named e. I wonder if * will preserve uniquness at all. If that were the case, then a*a' = e = a*a'' (if true) would imply that a' and a'' are the same.

.........
Okay, I guess I'm trying to interpret whatever the heck you were trying to write out with my little bit of linear algebra background. Totally lost on whatever you were saying with that terse solution.

 

[Kal]

Xeylode, when we say that the group is closed under *, we mean that the operation on two elements will stay within the group. So, if g_1 and g_2 are elements of the group, then g_1 * g_2 is also an element of the group.

Also, he was answering my question here.

 

[Hot_ArmS]

Can anyone think of how to do a flowchart of how to convert a binary number of any size and to a decimal number? For example, the response to an input 1111011 should be 123. You also can only add, subtract, multiply and divide in either base 2 or base 10, and can mix them.. You may use only one looping structure and one branch structure.

 

[Gates]

Going from binary to base ten notation and vice versa is a pretty common exercise in basic computer science classes. I'm surprised you didn't just find an algorithm someone else made by googling it and just use that. I personally don't know enough about programming to create such an algorithm but I found this, which may help:
http://stackoverflow.com/questions/4628036/binary-to-decimal-base-shift
Basically, it uses the character index to go through digits one at a time and multiply them by the corresponding powers of two (which themselves are kept track of incrementally) and add them to the total individually. You'd probably have to create an input variable and make the result more user friendly etc. but generally this is the right kind of code to use.

On an unrelated note, I really think we should have a link to Wolfram Alpha in the OP.

 

[PsychoIncarnate]

Supposedly Y=MX+B

But if I'm converting if from MX+Y=B I get all sorts of weird negatives and positives.

I wrote down rules for them, but I don't understand WHY some negatives and positives are switched during conversion and why some aren't/

IF: MX+Y=B THAN: Y=-MX+B
IF: -MX+Y=-B THAN: Y=MX-B
IF: -MX+Y=B THAN: Y=MX+B
IF: MX-Y=-B THAN Y=MX+B
IF: MX+Y=-B THAN: Y=-MX-B
IF: -MX-Y=-B THAN: Y=-MX+B
IF: -MX-Y=B THAN: y=-MX-B

Does anyone notice a pattern? Or an easier way to remember what to switch between negative and positive during conversion?

 

[Kal]

I don't really know what you're going for. If you're solving the general equation mx + y = b for y, you're naturally going to get different results for different values of m, b and y. But, it's sort of silly to try and memorize this, as "conversion" is more or less trivial, and you would be less likely to make an error if you just convert manually, rather than try to memorize the solution.

Anyway, one thing that might help is to consider the general solution:

if mx + y = b, then y = -mx + b

so, if you substitute (-m) for m in the above equation (i.e., if you try to convert -mx + y = b), you would get y = mx + b. All of your equations follow naturally in this way.

Sorry if that is not too helpful. I'm a little confused by your question.

 

[PsychoIncarnate]

I don't remember how to convert an equation. That's the problem.

 

[terr13]

Solve for y. If mx + y = b, then y= -mx +b. M and b can be negative, just remember two negatives is positive.

 

[Kal]

Yeah, you're just solving for y. If mx + y = b, you subtract mx from both sides to get y = -mx + b.

 

[PsychoIncarnate]

I figured out that if Y is positive, MX switches between positive and negative.

If Y is negative, MX keeps it's positive or negative but B switches between positive and negative.

You could also times the entire equation by -1 to make it into a positive.

Looking at all of them at the same time while studying late at night was confusing me for some reason

 

[terr13]

Okay, let me try to break it down a little bit.
Say you have Mx + y = b. This is the same as y = -Mx +b.

It doesn't matter whether or not y, m, x, or b are negative or positive, this formula is correct. Notice that if y is positive, then y = -Mx + b. If y is negative, but you want to solve for positive value of y, then we can multiply everything by -1, thereby getting -y = Mx -b. But -y will be a positive number, because y was negative to begin with.

 

[PsychoIncarnate]

No, I got it now. I was just confused before I realized that if Y is negative it's the exact opposite as if Y were positive.

I found an easy way to figure it out

 

[terr13]

What do you need to use this for? It seems like you don't quite understand it, but found a way that works without understanding it.

 

[PsychoIncarnate]

Math test. It's suppose to be a review chapter, but I haven't taken math in at least 6 years and probably have quite a bit of catching up to do.

I also don't remember doing it in high school, even though I was at this level in high school.

 

[terr13]

What kind of questions will be on the test related to this problem?

 

[PsychoIncarnate]

An example would be getting a slope and 2 points and trying to solve it in point-slope form or getting 2 points and a line equation

Getting X1 and Y1, and MX-Y=B

Converting it to Y=MX+B

Converting the slope (M) to the slope of point 2 (M2) by using M2=-1/M

Y-Y1=M2(X-X1)

 

[Hot_ArmS]

Getting X1 and Y1, and MX-Y=B

Converting it to Y=MX+B

just remember that just because it says y=mx + b doesnt mean that B has to be positive, B could be negative and would make it y=mx-b

Converting the slope (M) to the slope of point 2 (M2) by using M2=-1/M

sounds like an opposite reciprocal problem, so you mean how to find the equation of a line thats perpendicular to a given line?