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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]

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01 Feb 2006, 22:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 â€“ c = 3
VP
Joined: 29 Dec 2005
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01 Feb 2006, 23:33
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 â€“ 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.
Director
Joined: 17 Dec 2005
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Location: Germany
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02 Feb 2006, 01:41
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
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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02 Feb 2006, 01:45
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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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
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02 Feb 2006, 01:49
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 â€“ c = 3
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 22

Kudos [?]: 204 [0], given: 0

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02 Feb 2006, 02:08
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 . c can only be 1,2 and 3. 1 and 2 are already occupied. So c must be 3. Hence z = 9. SUFF.

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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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