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

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VP
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a b c + d e f -------- x y z If, in the addition problem [#permalink]

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08 Aug 2006, 11:07
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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

my question here is that it explicitly says that numbers ARE different, which is noted by different variables... if it didn't say so would it be correct to assume that some numbers could be equal?
VP
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08 Aug 2006, 11:21
First solving this one...

S1: 3a=f=6y
=> f = 6, y =1
=> c = 1 through 9
=> f +c = 7 through 15
=> z = 7, 8,9,0,1,2,3,4,5 Not sufficient.

S2: f-c = 3
c = 1 2 3 4 5 6
f = 4 5 6 7 8 9
------------------
z= 5 7 9 1 3 5

Not sufficient.

S1 & S2: f = 6, c = 3 z = 9
Sufficient.

Even if it is not implied that they are different single digits, due to the relationships given it would be difficult to come up with the same values for f/c/y/a.

Hoping I solved this right...
CEO
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08 Aug 2006, 12:12
St1:

y =1
f = 6
a = 2
c could be anything (3,4,5,7,8,9) Then f + c = (9,10,11,13,14,15) so z could be 9,3,4,5.: INSUFF

St2: f-c = 3
f - 4 5 6 7 8 9
c - 1 2 3 4 5 6
z = 5 7 9 1 3 5: INSUFF

Combined:
y = 1
f = 6
a = 2
c = 3

so z = 9: SUFF
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CEO
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08 Aug 2006, 12:14
Now try this one. I made a small change of sign.

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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VP
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08 Aug 2006, 13:15
VP
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08 Aug 2006, 13:18
Interesting..

S1: We get multiple values of z;
f = 6, y =1;
Not sufficient.

S2: f-c = 3

=> f = c+3

=> c = 0,1,2,3,4,5,6
=> f = 3, 4,5,5,6,7,8
--------------------------
=> z= 7, 7,7, 7,7,7,7 w/ a carry.

Sufficient..

ps_dahiya wrote:
Now try this one. I made a small change of sign.

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
VP
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08 Aug 2006, 13:25
Did we miss the case where a=f=y = 0 from S1?

S1 & S2:
Either a=f=y=0 or f=6, y=1, a=2 by S1.

But this would give us, c= -3 which cannot be the case.

But based on the case that f can have two values, maybe the answer is E.

u2lover wrote:
VP
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08 Aug 2006, 13:29
haas_mba07 wrote:
Did we miss the case where a=f=y = 0 from S1?

S1 & S2:
Either a=f=y=0 or f=6, y=1, a=2 by S1.

But this would give us, c= -3 which cannot be the case.

But based on the case that f can have two values, maybe the answer is E.

u2lover wrote:

nope zero case is not it... do you want me to tell you why or you want to figure it out on your own?
VP
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08 Aug 2006, 13:42
Can't say no to "figure out on my own" choice... can I ?

Let me try for ... 10 minutes and I'll ask if I can figure it out...

u2lover wrote:
haas_mba07 wrote:
Did we miss the case where a=f=y = 0 from S1?

S1 & S2:
Either a=f=y=0 or f=6, y=1, a=2 by S1.

But this would give us, c= -3 which cannot be the case.

But based on the case that f can have two values, maybe the answer is E.

u2lover wrote:

nope zero case is not it... do you want me to tell you why or you want to figure it out on your own?
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08 Aug 2006, 13:58
OK let me try again:

St1:
2 b c
d e 6
-----------
x 1 z

c can be 3,4,5,7,8, 9 and b+e must be 10 or 11
if (b+e) is 10 then c = 4,5,7,8,9
for (b+e) to be 10 possible cases are (7,3)
so now we have
2 7 9
d 3 6
-------
x 1 5

4 and 8 are remaing for d and x. This is not possible.
So b+e must be 11. For that c must be 3 and z is 9: SUFF

St2:
f-c = 3
a b c
d e f
------------
x y z
pairs of (c,f) are (1,4) (2,5) (3,6) (4,7) (5,8) (6,9)

I did all of these and only combination that is working is (3,6): SUFF

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VP
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08 Aug 2006, 14:01
I think this might be the reason...

Options = {0,1,2,3,4,5,6,7,8,9}

As each one is different, need to remove the number once it is used.

This might be leading to A... but I am having a hard time proving it.

Last edited by haas_mba07 on 08 Aug 2006, 14:13, edited 2 times in total.
VP
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08 Aug 2006, 14:04
Tricky one...

Good to see a solution...
Thanks PS/

ps_dahiya wrote:
OK let me try again:

St1:
2 b c
d e 6
-----------
x 1 z

c can be 3,4,5,7,8, 9 and b+e must be 10 or 11
if (b+e) is 10 then c = 4,5,7,8,9
for (b+e) to be 10 possible cases are (7,3)
so now we have
2 7 9
d 3 6
-------
x 1 5

4 and 8 are remaing for d and x. This is not possible.
So b+e must be 11. For that c must be 3 and z is 9: SUFF

St2:
f-c = 3
a b c
d e f
------------
x y z
pairs of (c,f) are (1,4) (2,5) (3,6) (4,7) (5,8) (6,9)

I did all of these and only combination that is working is (3,6): SUFF

VP
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08 Aug 2006, 14:12
ps_dahiya wrote:
OK let me try again

nope... but getting closer this is a tricky question I must say...
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08 Aug 2006, 14:24
u2lover wrote:
ps_dahiya wrote:
OK let me try again

nope... but getting closer this is a tricky question I must say...

OK.......two choices exhausted. If not these then must be A.
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VP
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08 Aug 2006, 22:28
anybody else want to try explain this question? i spent too much time on it, and really would like to know the shortest and most logical answer
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08 Aug 2006, 23:13
u2lover 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

my question here is that it explicitly says that numbers ARE different, which is noted by different variables... if it didn't say so would it be correct to assume that some numbers could be equal?

IMO, It's just A

All of those are single digit and '3a = f = 6y'
SO, y must be 1. otherwise f will be two digit number. then a=2
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09 Aug 2006, 08:38
OK lets see...

a b c
+
d e f
x y z

all variables are positive, distinct and single digits

Range of variables [1,2,3,4,5,6,7,8,9] 0 is not a positive number, so its not considered!

(1) 3(a)=f=6(y)

well...a has to be 2, no other solution works...so lets see
a=2, f=6, y=1

C cannot be 1, 2, or 6....so z cannot be 7, 2, or 8...

OK lets if y=1, then b and e cannot be {1,2,6},

i can calculate a few numbers and realize there is only 1 unique solution to this...which is sufficient to indicate (1) is sufficient..

(2) is in suff

A it is..
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09 Aug 2006, 08:49
jeunesis wrote:
All of those are single digit and '3a = f = 6y'
SO, y must be 1. otherwise f will be two digit number. then a=2

this is the key to this question!!!

OA is A
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09 Aug 2006, 08:57
u2lover wrote:
jeunesis wrote:
All of those are single digit and '3a = f = 6y'
SO, y must be 1. otherwise f will be two digit number. then a=2

this is the key to this question!!!

OA is A

Hmm, i still did not get it. i dont think the OE explains why the OA is A. I still think it is C.
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09 Aug 2006, 09:00
the thing is that in statement 1 the only possible value for y is 1, otherwise f will turn out to be two-digit number... if you put zero, then all numbers will equal to zero and that's not the case... cause they must differ
so if y=1, a=2 and we know that z-c=6... and only 1 pair of numbers satifies it z=9 and c=3... other pairs use 2 and 1, which were already used

I really broke my head over this one
09 Aug 2006, 09:00

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