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IMO C
1)numbers 13,38,63,88 are possible
NOT SUFF

2) all numbers 63+ 100n are possible where n>=0
NOT SUFF
when combined
only 63 is possible
SUFF

by tohellandback

Tue Sep 01, 2009 12:03 am
Forum: Problem Solving
Topic: Integers remainder
Replies: 2
Views: 1262

2^X- 2^(X-2)= 2^X- 2^X/4
so options 1 ,2 and 3 are out because x must be >13

check for the other two options

2^15- 2^15/4= 2^15-2^13=2^15(4-1)=3*2^13
answer D

by tohellandback

Mon Aug 31, 2009 11:59 pm
Forum: Problem Solving
Topic: value of x
Replies: 4
Views: 1217

first even number=2
last even number=n-1
total number of even numbers= (n-1-2)/2 + 1= n/2-3/2+1
=(n-1)/2
sum of A.P =total number/2(first term + last term)
in our case
(n-1)/4 * (2+n-1)=79*80
(n-1)(n+1)=4*79*80= 158*160
n=159

by tohellandback

Mon Aug 31, 2009 9:27 pm
Forum: Problem Solving
Topic: Maths Question
Replies: 5
Views: 2035

well,
1 says nothing about Y . thats why 1 is incorrect
for y=-100, x-y will be 200
for y=any other value it x-y will not be 200. Thats why NOT SUFF

2) is SUFF

by tohellandback

Mon Aug 31, 2009 8:55 pm
Forum: Data Sufficiency
Topic: x-y
Replies: 2
Views: 1753

IMO E

1) x>0
we need to know about Y
ex: x=10 Y=-8
(x-y)/(x+y) is 9
when x=10 y=-20
(x-y)/(x+y) is negative

NOT SUFF
2) Y<0
take the same cases as above
NOT SUFF
combined
again take the same cases as above

NOT SUFF

by tohellandback

Mon Aug 31, 2009 7:47 pm
Forum: Data Sufficiency
Topic: is (x-y)/(x+y)>1
Replies: 1
Views: 1183

It's basically asking how many 7's are there in 50!

answer is:
50/7 +50/49
= 7+1=8
answer A

by tohellandback

Mon Aug 31, 2009 7:35 pm
Forum: Problem Solving
Topic: faster way
Replies: 4
Views: 1311

there is a difference between

(a^b)^c and a^(b^c)
(a^b)^c will be a^(bc)
but for a^(b^c), you first need to calculate b^c because of the brackets.

and IMO OA given is right

by tohellandback

Mon Aug 31, 2009 7:25 pm
Forum: Problem Solving
Topic: 2^[(4-1)^2] / [2/^(3-2)]
Replies: 1
Views: 1562

Easy method: plugin numbers 16=8+8 1-false 2-false 3-true Answer D tricky method: 2^S=2^U + 2^V 2^S=2^V(2^(U-V)+1) 2^(S-V)= 2^(U-V) +1 now 2^(U-V) +1 will be ODD- when U>V or, 1+ (a fraction) when U<V but on the LHS, we have even when s>V or a fraction less than 1 when S<V so above cases are not pos...

by tohellandback

Mon Aug 31, 2009 7:09 pm
Forum: Problem Solving
Topic: tricky one
Replies: 2
Views: 1082

great score..congratulations..good luck with the apps..you seem to have an impressive profile...

by tohellandback

Sun Aug 30, 2009 8:00 pm
Forum: I just Beat The GMAT!
Topic: Done - 730 (Q51, V 37)
Replies: 5
Views: 2313

Re: combination sum - need clarification

Condition 1: it will be the equation X+Y+Z=5 and the answer is 4!/(2!*2! )--explanation for this result is discussed in some thread by Ian Stuart. If you don't get this..let me know isnt 4c2 = 6 though ur answer also sums upto 6 :) well you can always write my above expression as 4C2. and in fact I...

by tohellandback

Thu Aug 27, 2009 11:05 pm
Forum: Problem Solving
Topic: combination sum - need clarification
Replies: 8
Views: 1414

heshamelaziry wrote:
tohellandback wrote:2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

Could you explain this further? Thanks.
2*(2)=2^2
2^2+2^2= 2*2^2=2^3
2^3+2^3=2*2^3=2^4
2^4+2^4=2*2^4=2^5
2^5+2^5=2*2^5=2^6
2^6+2^6=2*2^6=2^7
2^7+2^7=2*2^7=2^8
2^8+2^8=2*2^8=2^9

by tohellandback

Thu Aug 27, 2009 10:24 pm
Forum: Problem Solving
Topic: Addin Exponents
Replies: 7
Views: 1751

2+2= 2*(2)=2^2
2^2+2^2= 2*2^2=2^3
and so on..
you will finally get 2^9

by tohellandback

Thu Aug 27, 2009 9:06 pm
Forum: Problem Solving
Topic: Addin Exponents
Replies: 7
Views: 1751

IMO E Stmt 1: Tens' digit of A = Tens' digit of B + Tens' digit of C Insufficient - This does not talk about the hundreth digit Case 1: B = 123, C = 321, A = 444 123 + 321 = 444 ------> This eqn satisfies the stmt 1 and also, from this we can conclude that the hundreth digit of A is equal to hundre...

by tohellandback

Thu Aug 27, 2009 8:58 pm
Forum: Data Sufficiency
Topic: Sum of digits
Replies: 13
Views: 1899

Thanks for the answers tohellandback, Can you please explain how you got 4!*4! in Qstn 4. Is there a formula for that kind of arrangement. Can you also tell me a place from where I can get Formula and lessons for permutation & Combinations. I am really bad at it... Thanks if the four boys are A...

by tohellandback

Thu Aug 27, 2009 8:43 pm
Forum: Problem Solving
Topic: Arrangements, permutations and combinations - interesting!!!
Replies: 12
Views: 5404