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VP
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abc + def = xyz If, in the addition problem above, a, b, c, [#permalink]
 01 Feb 2006, 22:28
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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VP
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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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Director
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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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CEO
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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
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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Senior Manager
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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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CEO
Joined: 20 Nov 2005
Posts: 2934
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 7
Kudos [?]:
37
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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  . 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.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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