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a b c + d e f -------- x y z If, in the addition problem [#permalink]
08 Aug 2006, 10:07

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 1 sessions

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?

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.

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

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

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 _________________

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

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