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abc + def = xyz If, in the addition problem above, a, b, c,

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VP
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abc + def = xyz If, in the addition problem above, a, b, c, [#permalink] New post 01 Feb 2006, 21:28
abc + def = xyz

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: Digits [#permalink] New post 01 Feb 2006, 22:33
lhotseface wrote:
abc + def = xyz
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


it is pretty obvious that it is between A and C. C works perfectly but A should also work. will work out more on it later. now i am tooooooooo tire and go to bed.
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 [#permalink] New post 02 Feb 2006, 00:41
from

1) we know that f=6, because y and a have to be integers and y=1/6 f and a=1/3 f

We yield

2bc + de6 = x1z

So

bc + e6 = 1z

c can't be
1,2,6, because we already have them
4,5, because with 6 z would be a digit that is already "used"
7,8,9 because c+6 would yield a number with two digits <19, but the second digit of xyz is 1 so that b and e both have to be 5, which isn't allowed

c=3
z=9


2) f-c=3

f=c+3

f can be everything <7

insuff

A
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 [#permalink] New post 02 Feb 2006, 00:45
I think C.
From main stem we can say that z is unit digit of (c + f)

St1: says that f = 6, y = 1 and a = 2.

So now we have 2bc + de6 = x1z

c+f i.e c+6 could be anything. INSUFF

St2: f-c = 3 means f+c = 3+2c. INSUFF

Combined:

From st1 we know f =6 and from st1 we have c = 3. So z will be 9. SUFF
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Re: Digits [#permalink] New post 02 Feb 2006, 00:49
A.


ABC
EFG
------
XYZ


From Stat 1, 3A=F=6Y,

since all are single digit number, Y is 1.
so Y=1, F=6, a=2

now lets put it in the addition equation

2BC
DE6
------
X1Z


or, 200+10B+C + 100D+10E+6 = 100X+10+Z


196 = 100(X-D) - 10(B+E) +(Z-C)

Therefore Z-C =6

possible value for C, 1,2,3. only 3 can be the vlaue of C because 1,2 are already value for Y and A respectively.


Therefore Z=9.


Stat 1. Suffiecient,

Stat 2 is not sufficient.


lhotseface wrote:
abc + def = xyz

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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 [#permalink] New post 02 Feb 2006, 01:08
ps_dahiya wrote:
I think C.
From main stem we can say that z is unit digit of (c + f)

St1: says that f = 6, y = 1 and a = 2.

So now we have 2bc + de6 = x1z

c+f i.e c+6 could be anything. INSUFF

St2: f-c = 3 means f+c = 3+2c. INSUFF

Combined:

From st1 we know f =6 and from st1 we have c = 3. So z will be 9. SUFF


I just missed because of the bold part :wall . c can only be 1,2 and 3. 1 and 2 are already occupied. So c must be 3. Hence z = 9. SUFF.

Answer should be A.
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  [#permalink] 02 Feb 2006, 01:08
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