vn_snoopy

Yes, at the time I post it, I also thought that my explanation is not as straightforward as yours :grad: , but I thought that it has something to look at :).

Hi, I just wanna explain a little more for Holden From the 5 digits 0,1,2,3,4,5, note that: 0 = 3 mod 0 1 = 3 mod 1 2 = 3 mod 2 3 = 3 mod 0 4 = 3 mod 1 5 = 3 mod 2 We need a five digits numbers divisible by 3. So, the sum of its digits must be divisible by 3. Now we look at the digits they give : 2 numbers which are divisible by 3, 2 others divide 3 remain 1, 2 others divide 3 remain 2. Now I call them mod0, mod1 and mod2. We see that mod1 + mod1 + mod2 + mod 2 -> mod0 which means that the sum of the four digits 1,2,4,5 are divisible by 3. mod1 + mod1 + mod2 + mod 2 + mod0 -> mod0 The number needed must have all these 4 digits, and another digit divisible by 3 (that is 0 or 3). Then, the number we get have digits We cannot substract any of these 4 digits, such as 1, and replace it by a number mod0, such as 0, since mod0 + mod1 + mod2 + mod2 -> mod2 and nothing change if we add the last digit mod0 to the number, the sum of its digits will still divide 3 remain 2. Now from the note above, we see that we have 2 options for digits : 1,2,4,5 and 0 1,2,4,5 and 3 For the first option, 1,2,4,5 and 0, we have 4 options for the 1st digit (except 0) and 4! option for the last 4 digits which gave us 4*4! = 96 options. For the 2nd option, 1,2,4,5 and 3, we simply have 5! = 120 options Totally, we get 96+120=216 I hope that it works.

Thanks Erin for your kind words :) . Last time I have to revise my writing some times before posting it :o . I need to practice much more to be able to post more here, and I will do it a little every day. I'm very happy that you also love Snoopy :tup: (What a loooveelyy dooogg... ). Also do I when I receive your reply :D .

Econ has posted the ans. The square root of r^6 equals the absolute value of r^3. Hence, we cannot know about the relation between colA and colB. For ex: Take r = 1: colA = colB = 1. Take r = -1: colA = 1, colB = -1.

Hi everyone, I've follow TestMagic for nearly one month. I've thought about having a post in the Introductory Forum many times. I firstly hesitated because I'm not good at writing. But, today :idea: , I do it . Erin, your site is great. I believe that it is better than the preparation courses for the TOEFL test I've attended. Moreover, it is FREE!!! Thanks so much Erin for your great sympathy to run the web site [goodjob] . I regret that I've found you only on week before my real TOELF test. I didn't do it well, especially the grammar section . I hope to pass it. I'm now preparing for the GRE. I'm really 'shocked' at the huge vocabulary for the verbal section, but I'll try to learn it as much as I can. Until now, I can just make some posts at the GRE Math forum. Since Maths is my major :grad:, I know it better than other things for the test. I have collected some really helpful advice on how to learn new words from old TestMagic members. Thanks so much for your kind help:tup: . I come from Viet Nam (the fist part of my id also tells it), and I love the little dog Snoopy:p . I hope that using its name for my nickname, as well as its picture as my avatar is not ... over the law. Good luck to all of you!!! :tup: vn_snoopy.

Whenever we rotate 90 degree, there always be some relation between sin and cos, and so the coordinates x and y. And, with a picture, u don’t have to memorize whether there is a minus or not. Anyway, i think that practice makes perfect... (that's what i'm trying to do ;) ) rectangulat coordinate rotation problem.doc

I think they are equal. The rectangular coordinate system is the plan with the usual coordinates Ox,Oy. The segment OP (sqrt(3), 1), going from the origin O(0,0) to the point P(sqrt(3),1), has the length of sqrt( sqrt(3)^2 + 1^2) = 2 and can be expressed as 2(sqrt(3)/2 , 1/2) = 2 (sin(pi/6) , cos(pi/6)). The angle between OP and the x axis is a = pi/6. Now when OP is rotated counterclockwise through an angle of 90 to position OQ, the angle increases 90 degree and the length remains the same. The angle b between OQ and the x axis is b = a + pi/2 = pi/6 + pi/2 The new x axis (of Q) = the length of OQ * sin(b) = 2sin(b) = 2sin(pi/6 + pi/2) = 2(-cos(pi/6)) = -1. We can do it more easily and more quickly with some drawing with which we can more easily see the relation between the old place and the new place of the segment.

one of my friend told that, since this is not a large number, we can remember it. And I have to admit that he has very good memory.

Until now I just use whatever come to my hand, usually the materials at university or some from the library. But they are quite old fashion. I also don't have the chance to study number theory at school and I'm looking for books on that. About the numerical analysis, I think that it would be a little out dated... Now I'm doing topology in the book of my teacher (in my native language). BTW, I live in South East Asean.

E is the ans since 2*6! = 1440. But I just simply do the computation.

I agree with you. We can only solve the problem if we know some more info about the length of PS' or about triangle S'SQ. I have added S' in the picture in the attachment.Real one_geometry-parallogram2.doc

Hi, I have a typo for my first post in the threat - I mean the case when both A and B hit. Sorry :) . I'm completely agree with Econ, I posted the first reply simply because I let the page so long without upload --> don't realized that u have answered :idea: .

Hi gurucool. I first thought as you that "exclusive" could also mean "mutually exclusive". But then P(x hits and y hits) = P(x hits) + P(y hits) = 2/3 + 4/7 > 1 -> impossible! BTW, my previous post is for locar... :)

You should add the case when both A and B fail.

What do they mean by "between"? Do they count the two numbers at the beginning and the end (which are 200 and 999) as "between" 200 and 999? If they do not, we have 798 numbers between 200 and 999, and the approximation percentage is also 12.5%. About the number required between 200 and 999: we have 2 choices for the hundreds digit (8 and 9), 5 choices for the units digit (the 5 odd numbers: 1,3,5,7,9), and 10 choices for the tens digit (0-9). Hence, we have 2*5*10 choices... as lmtuan has stated. ;)