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by cramya » Fri Jan 02, 2009 5:12 pm
Brent,
The question doesnt say different digits so I am assuming that's something u wanted in the question for a purpose.

Under that assumption, coming to the problem:

Stmt II

eb = 24

8,3 6,4
CANT BE 8,3 HAS TO BE 6,4

CARRY OVER WILL BE 1

SO A+D has to be 9

c and f have both got to be 0 since eb=6,4

we can find the sum of the digits(one single value 19)

SUFF

Stmt I

333
667

444
556

INSUFF

I would go with B)
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by Brent@GMATPrepNow » Fri Jan 02, 2009 5:26 pm
Stmt I

333
667

444
556

INSUFF
I'm not sure about your analysis of statement (1).
In both examples, the sum of the digits is 28
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by cramya » Fri Jan 02, 2009 5:37 pm
U r right; added it incorrectly. Got lucky with those numbers but since it said atleast 3 it opens up many possibilities so was trying to prove it as INSUFFICIENT. I have included a new set of numbers below

333
667

sum of digits = 28


444
556

sum of digits = 28

250
750

sum of digits = 19

Good problem! Whats the OA?
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by Brent@GMATPrepNow » Fri Jan 02, 2009 5:40 pm
Nice work - the answer is, indeed, B
Back to the drawing board :)
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by logitech » Fri Jan 02, 2009 7:01 pm
Cramya, as much as I like your number plugging methods, I findthem quite risky sometimes.

100(a+b) +10 (c+d) + (e+f) = 1000

There are three scenarios:

1 ) 900 + 100 + 0

2) 900 + 90 + 10

3) 1000 + 0 + 0


Statement 1) At least three numbers are bigger than 3

This eliminates the 3rd scenario.

but we still can have the 1st and 2nd scenarious - INSUF

a+b = 9, c+d =10 , e+f = 0 => 19
a+b = 9, c+d =9 , e+f = 10 => 28

Statement 2) cd=24

So it can be 8x3 or 6x4

8,3 does not work because

100 x 9 + 10 x 11 = 900+110 > 1000

So it needs to be 6,4 ---> which is the first scenario

So the answer is Beautiful question
LGTCH
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by cramya » Fri Jan 02, 2009 7:20 pm
Thanks Logtech! Like I always say we must always go with what we are most comfortable (conventional/unconventional) with, since that's the first line of attack we would be employing on the real exam.

Since I was tring to prove INSUFFCY of stmt I , I felt picking numbers was the best route to take at that time without spending too much time on the nitty gritties.

Also since I had done stmt II first all I had to show was the sum of the digits could vary with stmt I

For stm 1), I could have taken an example with 19 as the sum and then some random numbers that gives a different sum(satisfying atleast 3 condn).

Done...
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by logitech » Fri Jan 02, 2009 7:33 pm
Cramya,

I wish I am as comfortable as you are with plugging numbers since I always feel like I am missing out something. But I am reading your solutions very carefully to understand your methods better.

I am constantly in search of bullet proof solutions which we can apply under 2 minutes.

Keep up the good work master,
LGTCH
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by cramya » Fri Jan 02, 2009 7:37 pm
Logitech,

Thanks for the kind words! I always try to remember Ron's words of wisdom i.e. to not loose sight of whats in hand by trying to spend too much time on a textbook method(whats textbook for one may not be textbook method for the other...)

You too keep up the good work man! Some of the solutions u post are excellent too.

Good luck!

Regards,
CR
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