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plz post your reasoning can we use decimal numbers to

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plz post your reasoning can we use decimal numbers to [#permalink] New post 08 Aug 2006, 14:48
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A
B
C
D
E

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plz post your reasoning

can we use decimal numbers to disprove statement

Is A+B+C EVEN ?


ST1 A -C - B IS EVEN

ST2 ( A- C) B IS ODD


thanks
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 [#permalink] New post 08 Aug 2006, 15:08
e+e+e=e
o+o+o=o
e+o+o=e
o+e+e=o

(1) a-c-b=even, so either all even or there are 2 odds and in either case the subtraction is same as addition, so addition must be even
SUFF

(2) (a-c)b=odd, for odd result in multiplication, both factors must be odd..., so here (a-c)=odd and b must be odd, because there is no evens in multiplication with odd oucome... however, (a-c)...either a or c could be odd, so we will get o+o+e, which makes outcome even...
SUFF

so I say D... let's see if I screwed up
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 [#permalink] New post 08 Aug 2006, 15:26
mandy,
If it does not say numbers must be integers, then yes, you can try using decimals or fractions.

Anyway,

I am getting (C)

ST1. A-C-B is even, or A-C-B=2k

If we choose A,B,C all integers and even, then A+B+C is even
However, if we use fractions, it does A+B+C does not have to be even

Let's see: A=2.2, B=0.1, C=0.1 and A-B-C=2 even. But A+B+C is not

ST2. (A-C)*B is odd

It follows that either
(1) A-C is even and B is odd
or (2) A-C is odd and B is even
or (3) A-C is a fraction and B is integer such that (A-C)*B=odd
or (4) A-C is integer and B is a fraction such that (A-C)*B=odd

Not sufficient.

Now, combine ST1 and ST2.

Neither A-C nor B can be fractions, otherwise ST1 is not valid.

So, all numbers -- A, B, and C -- must be integers.

Moreover, either 1 or all 3 must be odd (otherwise ST2 is violated.)

Therefore, A+B+C is even

(C) is sufficient
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Re: DS even [#permalink] New post 08 Aug 2006, 17:35
mand-y wrote:
can we use decimal numbers to disprove statement?
Is A+B+C EVEN ?

ST1. A -C - B IS EVEN
ST2. ( A- C) B IS ODD



E. good question. if A, B and C are not whole numbers/integers or are fractions, we cannot say.

initially i also thought B but after going through your question, should be E.

this is indeed another good question............................
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 [#permalink] New post 08 Aug 2006, 18:51
Prof,

Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?

I am still convinced that it should be (C)...
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 [#permalink] New post 08 Aug 2006, 19:05
v1rok wrote:
Prof, Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?

I am still convinced that it should be (C)...

if a = 4.5, b = 1 and c = 1.5
st 1. a - c - b = 4.5 - 1.5 - 1 = 2.........even
st 2. (a - c) b = (4.5 - 1.5) 1 = 3........odd

a + b + c = 4.5 + 1.5 + 1 = 7 ...odd.

if a = 4, b = 1 and c = 1
st 1. a - c - b = 4 - 1 - 1 = 2.. even
st 2. (a - c) b = (4 - 1) 1 = 3 .. odd

a + b + c = 4 + 1 + 1 = 6 ... even.

Quote:
Is A+B+C EVEN ?

ST1. A -C - B IS EVEN
ST2. ( A- C) B IS ODD
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 [#permalink] New post 08 Aug 2006, 22:53
This is E.

NEVER assume integers until explicitly stated.
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 [#permalink] New post 09 Aug 2006, 17:39
THANKS FOR yours answers but i thought that even number have to be integer
becaause even is an integer when divided by 2
oa will come
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 [#permalink] New post 09 Aug 2006, 17:41
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.


Good ANSWER oa is E
plz can you explain me why
thanks
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 [#permalink] New post 09 Aug 2006, 18:15
consider AD/BCE,
A---
let A be 5.1,C be 2.1,A be 1
A-B-C=5.1-1-2.1=2(even)
but A+B+C=8.2 not even...
but if A is 6,B is 1 C is 3 both stmts are even hence A and D goes out
now consider B,
noe (5.1-2.1)1=3,odd
consider the same above combinations above,we will get yes for one
combinaton and no for another combination.hence B goes out
similarly for c we will get yes for 6,1,3 while no for 5.1,2.1,1
hence C also goes out
and giving us the answer E
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 [#permalink] New post 09 Aug 2006, 20:14
mand-y wrote:
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.


Good ANSWER oa is E
plz can you explain me why
thanks


A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E.

Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED
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 [#permalink] New post 10 Aug 2006, 07:15
ps_dahiya wrote:
mand-y wrote:
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.


Good ANSWER oa is E
plz can you explain me why
thanks


A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E.

Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED

so I understood well the whole sum is an even integer it does not mean
that each part -each - has to be an integer
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  [#permalink] 10 Aug 2006, 07:15
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plz post your reasoning can we use decimal numbers to

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