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a b c d e f ----- x y z If, in the addition problem above,

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a b c d e f ----- x y z If, in the addition problem above, [#permalink] New post 05 May 2006, 21:43
a b c
d e f
-----
x y z

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y

(2) f – c = 3
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Re: Addition of variables! [#permalink] New post 21 May 2006, 06:31
chiragr wrote:
a b c
d e f
-----
x y z

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y
(2) f – c = 3


go with D. the only values that work for c and f are 3, and 6 respectively...
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 [#permalink] New post 21 May 2006, 07:20
with (1) we know f=6

from (2) we know c=3, so C looks good.

Prof, D is quite possible if we put in more time on figuring out what the other #s can be.. but can you explain the steps how you got it or did you just make an intellectual guess ( :) which we know you are very good at)
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Re: Addition of variables! [#permalink] New post 21 May 2006, 07:39
Professor wrote:
chiragr wrote:
a b c
d e f
-----
x y z

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y
(2) f – c = 3


go with D. the only values that work for c and f are 3, and 6 respectively...


D it is....anyhow prof. c and f can be 2 and 5 regarding only (B)....z is same though...
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 [#permalink] New post 21 May 2006, 08:08
I go with C.

Statement 1: From this we get f =6, a=2, y=1 . Not sufficient for z.

Statement 2: f=c+3. This can give a few values.

Combining these, we get z = 9
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 [#permalink] New post 21 May 2006, 11:39
remgeo wrote:
I go with C.

Statement 1: From this we get f =6, a=2, y=1 . Not sufficient for z.

Statement 2: f=c+3. This can give a few values.

Combining these, we get z = 9


hm..Thanx....Statement 1 is not sufficient
but Statement 2 is sufficient.
cuz z is positive single digits, (f,c) can be
(4,1) (5,2) (6,3) (7,4) (8,5) (9,6)

so c-f = 7 always

So, B
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 [#permalink] New post 21 May 2006, 11:40
oops....

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits,

it's addition problem. sorry it's C
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 [#permalink] New post 21 May 2006, 18:35
did you guys, put the diffent digit values for other variables too?

if you go after only c, f and z, yes it is C but if go after all values, only 3, 6, and 9 work for c, f and z.

in both circumstances, z is 9.
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 [#permalink] New post 21 May 2006, 18:42
Professor wrote:
did you guys, put the diffent digit values for other variables too?

if you go after only c, f and z, yes it is C but if go after all values, only 3, 6, and 9 work for c, f and z.

in both circumstances, z is 9.


Can you please explain prof, how can you deduce the answers from either 1 or 2?

I honestly can't think of any answer other than C.
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 [#permalink] New post 21 May 2006, 18:54
St2:
f-c = 3
Can be (f,c) = (3,0), (4,1), (5,2).... (9,6)
Insufficient.

St1:
Since all the values are single digits, y must be 1. If y = 2, f will be a double digit number. But we have no information abotu c, so we cannot derive z. Insufficient.

Using St1 and St2:
we know f = 6, c = 3 and z = 9. Sufficient.

Ans C
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 [#permalink] New post 21 May 2006, 19:27
kapslock wrote:
Professor wrote:
did you guys, put the diffent digit values for other variables too?
if you go after only c, f and z, yes it is C but if go after all values, only 3, 6, and 9 work for c, f and z.

in both circumstances, z is 9.

Can you please explain prof, how can you deduce the answers from either 1 or 2?
I honestly can't think of any answer other than C.

how could you fill the values for all variables?

Quote:
a b c
d e f
-----
x y z

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y
(2) f – c = 3


from i, y = 1, a = 2, f = 6. lets put these values:

2 b c
d e 6
-----
x 1 z

the following is the only way to have the different single digit values for these variables. however, the values for some variables can be changed, the value of z remains 9.

2 7 3
5 4 6
-----
8 1 9
hope this helps....

same applies with ii.
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 [#permalink] New post 21 May 2006, 19:31
Professor wrote:
kapslock wrote:
Professor wrote:
did you guys, put the diffent digit values for other variables too?
if you go after only c, f and z, yes it is C but if go after all values, only 3, 6, and 9 work for c, f and z.

in both circumstances, z is 9.

Can you please explain prof, how can you deduce the answers from either 1 or 2?
I honestly can't think of any answer other than C.

how could you fill the values for all variables?

Quote:
a b c
d e f
-----
x y z

If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y
(2) f – c = 3


from i, y = 1, a = 2, f = 6. lets put these values:

2 b c
d e 6
-----
x 1 z

the following is the only way to have the different single digit values for these variables. however, the values for some variables can be changed, the value of z remains 9.

2 7 3
5 4 6
-----
8 1 9
hope this helps....

same applies with ii.


2+6, 1+6, 0+6, 3+6 all end up with single digits for z.
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 [#permalink] New post 21 May 2006, 19:32
ywilfred wrote:
St2:
f-c = 3
Can be (f,c) = (3,0), (4,1), (5,2).... (9,6)
Insufficient.

St1:
Since all the values are single digits, y must be 1. If y = 2, f will be a double digit number. But we have no information abotu c, so we cannot derive z. Insufficient.

Using St1 and St2:
we know f = 6, c = 3 and z = 9. Sufficient.

Ans C


0 is not +ve digit.
you guys are considering only c, f and z but not other variables.

the question is "how to assign the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 to the variables a, b, c, d, e, f, x, y and z?
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 [#permalink] New post 21 May 2006, 20:06
missed out the positive part and the different digits part... :oops:
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 [#permalink] New post 21 May 2006, 21:35
St2:
f-c = 3
Can be (f,c) = (9,6) (8,5) (7,4) (6,4) (5,2) (4,1)

(8,5) (7,4) (6,4) (5,2) (4,1) --> all are out as this will not allow a, b, c, d, e, f, x, y, and z to be represented by unique single digits


Sufficient.

St1:
We know y = 1, f = 6, then a = 2. Now we have

2 b c
d e 6
-----
x y z

If c is some other value other than 3, then b,e,d,x, and y cannot be represented by unique single digits.

E.g If C = 1,

Then we have

2 b 1
d e 6
-----
x y 7

So we have 3,4,5,8,9

We know 3 cannot be added to 4 as 7 is already used for z. But we can add 3+5 to get 8.

2 3 1
d 5 6
-----
x 8 7

Now 4,9 are left.

2+4 cannot be equal to 9 and 9+2 cannot be equal to 4 and is not valid as it results in a double digit number.


Ans D
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 [#permalink] New post 22 May 2006, 07:54
One more for D

3a = f
2a=6y
f=6y

so we know y =1, a = 2, f =6

since we are told they are single DISTINCT positive digits, we know C cannot be 1 or 2, the only other thing that works is C=3. So I is sufficient

II. f-c=3

same logic
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 [#permalink] New post 22 May 2006, 10:15
Prof, perfect explanation.
  [#permalink] 22 May 2006, 10:15
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