thankont

If 5 persons are arranged on a line then we have 5*4*3*2*1 or 5! possible arrangements. But what if people are seated on a circular table? Actually the only difference here is that arrangements A,B,C,D,E and B,C,D,E,A and C,D,E,A,B and D,E,A,B,C, finally E,A,B,C,D are the same (relative positions are the same). So divide 5! by 5 which equals 4! = 24

The only combination which gives a difference of squares is (x-y)(x+y). So out of 6 combinations (why 6: call A=x+y and in a similar fashion B,C,D for the rest,.. so you have 6 combinations AB, AC, AD, BC, BD, CD or 4C2 using the C symbol for combinations or ( ) parentheses, depending of the book you are reading)

p= 1/(4C2) =1/6 since for numerator we have only x^2-y^2 and for denominator from 4 subjects choose 2 (combination)

Common factor in our case is 4^14. From properties of exponents 4^17=4^14*4^3 (same base add powers for multiplication)

A fast approach (ackowledging that Brent's approach with Lav's solution is the appropriate way to go) is the following: a^2 = b^2+5^2+5^4+5^6 (1) so if we try "clean" multiples of 5 so as to achieve a perfect square while b=5 does not work however if b=25 then (1) becomes 5^2*(1+5^2)^2 which is a perfect square.

4^17-2^28 = 4^17 - 4^14 = 4^14(4^3 - 1) = 4^14 * 63 = 7*... so largest prime factor is 7

Look try another strategy. Start working with the OG and spot your weak points. I know there is an excel sheet on the net that records your mistakes. After that, just concentrate on those sections where you don't feel comfortable. Also if and when you retake the test always write something down immediately. You lose valuable time just looking at the question. After all, we all make mistakes and it depends on ourselves to try and minimize them. Don't forget to take the gmatprep tests before giving the original one. You will have a strong indication if you are ready to proceed. All the best and keep it up...

4C1(RWWW)=4*1/4*2/3*1/2*1 need 4C1 since right one -R- could be in any of the four positions. If it is on the first position then prob. of putting the correct one there is 1/4, remaining ones are 3, so if we want second envelope to be the wrong one then p=2/3 etc.

Since I couldn't find the solution I tried it out, so 13x10^5=1,300,000 (1) and 2x13x10^4=260,000 (2) and (1)-(2)=1,040,000 (3) one more step 3x13x10^3=39,000 (4) so (3)-(4)=1,001,000 but we want a 1 at the unit digit so add 1 only does not work (not a multiple of 13) now try to add 1001 and this works since 1001/13=77 so our number is 1,002,001

given n=2,5,8,11,14,17,... t=3,8,13,18,23,... i) n=2,7,12,17,22,... ii)t=3,6,9,12,15,18,... so first common terms for n,t are 2,3 so right away remainder is 6 next common terms for n,t are 17,18 and 17x18=306 remainder 6

Q3) 8c1+8c2+8c3 = 8+28+56=92

if you plug in numbers you will see it right away. if n=4 then (n+1)(n+2)=5*6 is a multiple of 3, if n=5 then we have (n+1)(n+2)=6*7 again multiple of 3 but if n=6 then (n+1)(n+2)=7*8 not a multiple of 3.

x=-b/a so if x>0 then b,a should have opposite signs 1) insuff 2) insuff combining (adding) we only get a>0 (no hint about b) so insuff. (we cannot subtract since inequality reverses)

say if n=3 then (n+1)(n+2) or 4*5 is not a multiple of 3...

line y=3x+2 contains point (r,s) or in other words 3r+2-s=0 (want to show) 1) insuff. 2) insuff. 1)-2) = 15*(3r+2-s)=0 so 3r+2-s=0

imo p=2/3 (since product is not a multiple of 3 if n is a multiple of 3 and there are 33 multiples)

it is B) since from statement 2) L.H.S is either +ve or 0 and R.H.S is either -ve or 0. So since a +ve number is not less than a -ve number, we look only at the equal sign so |x-3|=0 or x=3

I like this one: Permutation and combination tutorial

look at first line we could have: 5 -> 1, 4 ->2, 3->3, 2->4 or 10 rectangles in total x 5 lines = 50 x 2 for columns so up to know we have 100 rectangles. look at two lines: 5->1, 4->2, 3->3 or 6 rectangles in total x 4 = 24x2 for columns =48 same with 3 lines: 5->1,4->2, or 3 rectangles x 3=9x2 for columns =18 add 4 for the rest total 170

voting for 32 also: 8*6*4/3! =32 or 8C1*6C1*4C1/3! (choose from the 8 persons 1 take out the married one then from 6 choose 1 and finally from the rest 4 choose 1 again and since ABC,ACB,BCA are the same here, divide throughout by 3!)

If x=y then k=15 and x/(x+y), y(x+y) both x then k>15 (tilted to the right)

x=5, y=2

from 1) x is not 0 or 7 and from 2) x is not 5 or -3 so in order for x(x-5)(x+2) = 0 we must have x=-2 or x

|x| = x if x>0 or -x if x1 or -x+1 if x so if 0

Yes, this is what I got...