donlau

for 2) b+c+d=300.....(1) a =>a+c+d (1)+(2) a+b+2(c+d) 200+2(c+d) 2(c+d) c+d c+d

when a.b=0 and b=1 L.S.=0 R.S.=a+1-a=1 LSRS, impossible. b cannot be 1.

1)a^2>=0 2)b0 =>b^2>0 1)+2) a^2+b^2>0

589. Please reserve 700 for me and for my GMAT!

Interesting! See the second post from Agent_777.

Vreddy, Just want to share some idea. it seems A is sufficient. I will do it by tomorrow if possible.:)

Zorrillo, That is a good question. I am still comtemplating on a date. I am mulling to take both GRE and GMAT. Do you have any GRE info to share? I would appreciate that.

when b >=8 |b+8|+|b-2| = 2b+6, this is always greater than 10. Sufficient So, answer is B.

Vreddy, You mean A? Sufficient. A is the answer?

For GMAT, http://www.mba.com/mba/TaketheGMAT/ToolsToHelpYouPrepare/GMATPrepProducts/PowerprepSoftware.htm For GRE, http://www.gre.org/pprepdwnld.html

This quotation is from John D. Rockefeller, 1839-1937, American Industrialist, Philanthropist, Founder Exxon.

For your reference, another type was done by Vreddy before. http://www.TestMagic.com/forum/topic.asp?TOPIC_ID=10873

101010x2^-4 therefore, a),b) and c) 1.999999 close2 2 .999999 close2 1 .999998 close2 1 (1.999999)(.999999)+(.999998)(.999999) close2 2x1+1x1=3

Markus, you are welcome.:)

Hi Markus, Locate MATHREV1.HLP, MATHREV2.HLP, MATHREV3.HLP and MATHREV4.HLP from c:etsppgmatppgmat first and then use some tools such as "Windows HLP To RTF2.6" to convert windows help file to one common document file.

Let f be a function satisfying f(xy)=f(x)/y for all positive real number x and y. If f(500)=3, what is the value of f(600)? (A)1 (B)2 ©5/2 (D)3 (E)18/5

In general, the winner is the one who can make the sticks multiple of 6 on the table first and maintain them to be multiple of 6 to another player in the rest of the game. Therefore, another player has to pick 1,2,3,4or5 sticks from 6 sticks on the table in the turn before the last turn. In the last turn, the winner will pick the rest of sticks-5,4,3,2or1 on the table. This implies that the loser will be the one who picks the sticks multiple of 6 on the table such as 6, 12, 18, 24 .. This is not a fair game.

I hope I do not misunderstand the question. when 6 sticks are on the table, no matter how many sticks the current player takes from 1 to 5, the rest of them on the table can be taken by another player. Let us play the game and there are 11 sticks on the table. Now I take 5 sticks and then there are 6 sticks on the table. Let us see who is the winner.

Other than 1,2,3,4or5, 6 is the lucky number to maintain the control so that another player won't win in the next round. Among 5 chooice, 12 is the first number that the current player loses the control of the game. For 7, take 1 to make 6 sticks. For 10, take 4 to make 6 sticks. for 11, take 5 to make 6 sticks.

Vreddy, don't bother you? I would like to join you. :) 88o88 o4o4o oo oo o4o4o 88o88 we can observe the relationship since the frame has 1 inch side, M=p+4 M+52=M+(M+8)=>M=44 Therefore, 44=p+4=>p=40

This is another way. Put the room in Coordinate Geometry. Coord. of one corner at the bottom = (0,0,0) Coord. of the bulb = (6,8,10) or (8,6,10) The distance btw both = sqrt(6^2+8^2+10^2)=sqrt(200)=10*sqrt(2)

For 1a, it implies #soln of non-negative integers for x1+x2+x3+x4=10 ans=C(10+4-1,10)=286 For 1b, 10 diamonds as distinct and their order within each shelf are important or their arrangement within each shelf is irrelevant? For irrelevant, this can be done in 4x4x..x4=4^10ways

Bazooka76, thanks. Zorrillo, I can see no new topics lately. So donlau for problem solving and ron for asking questions. I have no intention to do harm. See you all around in math forum.:)

BTW, Vreddy, I use another nick lately This is ron. I have seen that you gave a comment yesterday. Before that, I found that Rumen just took a look and left. Anyway, Zorrillo, Congratulation![banana] Well done on your GRE. I don't like to join the party.:)

When last digit is 0 total numbers are 8x9x1 when last digit is 2,4,6 or8 total numbers are: 1) 8x1x1 when middle digit is 0 or 2) 7x8x1 when middle digit is not 0 Total for 2,4,6 and 8 are 4x(8x1x1+7x8x1)=4x64=256 Hence, total for all = 8x9x1+256=328 amendment: The following is another simple counting approach: 1. can be combined with 2. 8x1x1+7x8x1=8x8x1 this implies there are 4 ways to put {2,4,6,8} into final digit, 8 ways to 1st digit without 0 and 8 ways to 2nd digit with 0. #ofways=8x8x4 Hence, total for 0,2,4,6 and 8 = 8x9x1+8x8x4=328