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I'm going to make a sticky thread where the very very basic

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 [#permalink] New post 28 May 2006, 03:48
Thanks a lot everyone. I need massive amounts of help with my math, so this will be extremely beneficial. This is my starting point in the quest for 700!!
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 [#permalink] New post 06 Jun 2006, 06:52
HongHu and others,
Thank you very much for putting this together. I just spent three hours going over the 9 pages meticulously and believe me there were things I didn't know I didn't know.
Hope this will help me get to my intended score.

Thanks again.
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 [#permalink] New post 09 Jun 2006, 18:09
Very good post ..

Thanks
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 [#permalink] New post 30 Jun 2006, 11:52
ywilfred wrote:

Operations involving the same bases:
- Keep the base, add or subtract the exponent (add for multiplication, subtract for division)

E.g. Same base - multiplication

(5^3)(5^6) = 5^(3+6) = 5^9

E.g. Same base - division

(6^3)(6^2) = 6^1 = 6


Great post but one error-- in the last example, it should read: (6^3)/(6^2) = 6. As it is posted ( ((6^3)(6^2) ) the answer would actually be 6^5.
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 [#permalink] New post 30 Jun 2006, 13:46
HongHu wrote:
Another example:
|x+4|>4
If x+4>=0, then x+4>4. Solve for both you get x>=-4, x>0. So your solution is x>0.
If x+4<0, then -(x+4)>4, ie. x+4<-4. Solve for both you get x<-4, x<-8. So your solution is x<-8.
You final solution is x>0 or x<-8.


HongHu, for the first part, why is the solution x>0 and not x>=-4 and x>0?
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 [#permalink] New post 03 Jul 2006, 12:23
You need to find the intersection of the two sets x>=-4 and x>0, which turns out to be x>0. If you could draw it on a number line, it will be easier to see. It's like this (please excuse the crappiness of the graph ;)):

Code:
    -------------------
    |   ---------------
    |   |
----------------------------
   -4   0


You can see the overlapping part is x>0.
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Re: Working with Ratios [#permalink] New post 19 Sep 2006, 13:26
HongHu wrote:
Working with Ratios

Example 2:

In a picnic 60% people ate two hotdogs, 30% people ate one hamburger, and 10% people ate one hotdog. The total number of hotdog and hamburgers consumed is 80. How many hamburgers and hotdogs are consumed?

People:
T:H:O=6:3:1 (1)
Food:
2T+H+O=80
From (1)
T:H:O:(2T+H+O)=6:3:1:16
Therefore
T:(2T+H+O)=3:8=30:80 30 people ate two hotdogs
H:T=3:6=1:2=15:30 15 people ate one hamburger
O:T=1:6=5:30 5 people ate one hotdogs
Total people 50, total hotdogs 65, total hamburgers 15.
Verify, total hotdogs and hamburgers=65+15=80.


I'm confused guys. Total Hamburgers = 15? Is that 30% of 80?

Can we instead say: 30% of 80 = 0.3 * 80 = 24 hamburgers used
80 - 24 = 56 Hotdogs used.

What am I missing?
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 [#permalink] New post 09 Oct 2006, 09:26
You need to make sure which figures are refering to the people and which are refering to the food. Since there are people who eat two hot dogs, number of foods are more than number of people. So 30% people eat hamburger doesn't mean hamburgers are 30% of the total food.
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 [#permalink] New post 12 Oct 2006, 03:51
I wnet through all the postings of this thread and was able to refresh a lot my forgotten concepts.
I am new to this and my sincere thanks to all.
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 [#permalink] New post 19 Oct 2006, 18:34
I think if the last three digits of a number are divisible by 8, then the number is divisible by 8.
does anyone disagree?

ywilfred wrote:
I'll add on some more over a period of time, but here are for starters:

Divisibility rules: - Know them well

Integer is divisible by:
2 - Even integer
3 - Sum of digits are divisible by 3
4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4
5 - Last digit is 0 or 5
6 - Integer is divisbile by 2 AND 3
8 - Integer is divisible by 2 three times
9 - Sum of digits is divisible by 9
10 - Last digit is 0

Some other things to note:
- If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.

(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))

- Remember to include '1' if you're asked to count the number of factors a number has
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Great work, everyone. Thanks very much! [#permalink] New post 21 Oct 2006, 21:05
Hi everyone,

Great posts and absolutely wonderful work! Greatly appreciate all your insights.

Thanks! :-D
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 [#permalink] New post 22 Oct 2006, 10:17
chaiku wrote:
I think if the last three digits of a number are divisible by 8, then the number is divisible by 8.
does anyone disagree?


That's correct. Multiples of 1000 is divisible by 8 so all you need to worry about is the last three digits. Similarly you only need to look at the last two digits when you want to see if it is divisible by 4.
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 [#permalink] New post 25 Oct 2006, 10:26
ywilfred wrote:
adding more...

Interesting properties:
- Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)

E.g. Adding 3,4,5,6 will give a sum with 4 as a factor
Adding 2,3,4 will give a sum with 3 as a factor

etc


3+4+5+6=18
4 is a factor of 18?????

2+3+4+5=14
4 is a factor of 14?????
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 [#permalink] New post 04 Nov 2006, 10:13
I have posted my math formula sheet in my thread at:
http://www.gmatclub.com/phpbb/viewtopic.php?t=37592

Hope this helps.
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 [#permalink] New post 29 Nov 2006, 00:54
banerjeea_98 wrote:
Dan wrote:
If x > y, then x - y divides x^n - y^n.


interesting....can someone provide a proof for this...thx


The root of any polynomial of the form ax^n + bx^(n-1) + ..C = 0 is basically that value of x that makes the left hand side 0.

To take an example a simple polynomial

x^2 + 2x + 1 = 0

x + 1 is a root of the polynomial.

Why? Because when you plug x = -1 in the eqn you get 0.

Thus to extend the analogy x^n - y^n will always be zero when x = y. In other words x-y will be a root of the polynomial.

QED!
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Geometry [#permalink] New post 11 Dec 2006, 06:06
There have always been some confusions about a basic principle in geometry:

Special shapes are always instances of the group that it belongs. For example, a square is a polygon with four equal sides, four right angles, and parallel opposite sides. Therefore, a square is a special case of a regular quadrilateral, rectangle, rhombus, parallelogram, and isosceles trapezoid.
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square roots [#permalink] New post 22 Dec 2006, 13:22
According to mathworld, a square root of x is a number r such that r^2=x. In other words, any positive real number has two square roots, one positive, sqrt(x), and one negative, -sqrt(x). Although in common usage, square root is generally taken to mean the principle square root, i.e. sqrt(x).
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Last edited by HongHu on 27 Apr 2008, 11:17, edited 2 times in total.
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Re: square roots [#permalink] New post 23 Dec 2006, 01:41
HongHu wrote:
According to mathworld, a square root of x is a number r such that r^2x. In other words, any positive real number has two square roots, one positive and one negative. Although in common usage, square root is generally taken to mean the principle square root, i.e. sqrt(x).


Sorry.... I must disagree. The function is not reciprocal on the negative values.

Sqrt() always returns a positive number by definition. What is matter is that x must be positive in the sqrt() in order to make the equation existing (at least in the real number set of function).

* Sqrt(4) = 2 cannot = -2.
* Sqrt(9) = 3 cannot = -3.

As well, if we define x = 4, we have sqrt(x) = sqrt(4) = 2.

I join a print of the 2 functions. As we observe, for the power of 2, 2 values of x are possible for 1 value of y. But, for the square root, 1 unique value of x match 1 value of y and moreover, x must be positive.

:)
Attachments

Fig1_Square root function.GIF
Fig1_Square root function.GIF [ 2.42 KiB | Viewed 2998 times ]

Fig2_power 2 (reciprocal).GIF
Fig2_power 2 (reciprocal).GIF [ 2.4 KiB | Viewed 2942 times ]

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 [#permalink] New post 23 Dec 2006, 10:51
I agree that sqrt(x) is always positive. However it is also true that x=r^2 have two square roots, one is sqrt(x) and the other is -sqrt(x). I've edited the previous post to make it more clear.
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 [#permalink] New post 22 Jan 2007, 11:55
x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.

I know this is basic but cannot see how you`re getting from : if x<0 and x^2-1<0 to x>-1. I get -x>1. Its been a while since I`ve done this stuff !
CHeers for the help. :?
  [#permalink] 22 Jan 2007, 11:55
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