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a b c d e f  x y z If, in the addition problem above, [#permalink]
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05 May 2006, 22:43
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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 â€“ c = 3



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Re: Addition of variables! [#permalink]
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21 May 2006, 07:31
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 â€“ c = 3
go with D. the only values that work for c and f are 3, and 6 respectively...



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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]
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21 May 2006, 08:39
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 â€“ 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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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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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 cf = 7 always
So, B
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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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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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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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St2:
fc = 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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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 â€“ 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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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 â€“ 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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ywilfred wrote: St2: fc = 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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missed out the positive part and the different digits part...



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St2:
fc = 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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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. fc=3
same logic



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Prof, perfect explanation.










