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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink ]
18 Aug 2008, 09:21

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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 – c = 3

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Re: DS - addition problem [#permalink ]
18 Aug 2008, 09:26

sarzan 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 – c = 3

C

Need to know c and f.

(1) Insufficient

y = 1 (anything greater than that, and f is a double digit number)

f = 6

a = 2

(2) Insufficient

f could 3-9, c could be 0-6.

(1) and (2) Sufficient

f = 6

6 - c = 3

c = 3

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Re: DS - addition problem [#permalink ]
18 Aug 2008, 09:42

sarzan wrote:

zoinnk wrote:

sarzan 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 – c = 3

C

Need to know c and f.

(1) Insufficient

y = 1 (anything greater than that, and f is a double digit number)

f = 6

a = 2

(2) Insufficient

f could 3-9, c could be 0-6.

(1) and (2) Sufficient

f = 6

6 - c = 3

c = 3

No, thats not it.

Sorry, I didn't read the question carefully. I didn't see that the numbers are distinct.

A

In that case:

2 b c

d e 6

-----

x 1 z

The only possibilities that work are:

2 7 3

5 4 6

-----

8 1 9

or

2 4 3

5 7 6

-----

8 1 9

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Re: DS - addition problem [#permalink ]
18 Aug 2008, 09:53

[/quote] Sorry, I didn't read the question carefully. I didn't see that the numbers are distinct. A In that case: 2 b c d e 6 ----- x 1 z The only possibilities that work are: 2 7 3 5 4 6 ----- 8 1 9 or 2 4 3 5 7 6 ----- 8 1 9[/quote] But why did you pick y=1?

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Re: DS - addition problem [#permalink ]
18 Aug 2008, 10:23

sarzan wrote:

Sorry, I didn't read the question carefully. I didn't see that the numbers are distinct. A In that case: 2 b c d e 6 ----- x 1 z The only possibilities that work are: 2 7 3 5 4 6 ----- 8 1 9 or 2 4 3 5 7 6 ----- 8 1 9 But why did you pick y=1?

I explained that in my first post:

y = 1 (anything greater than that, and f is a double digit number)

f = 6

a = 2

Last edited by

zonk on 18 Aug 2008, 11:42, edited 2 times in total.

Re: DS - addition problem
[#permalink ]
18 Aug 2008, 10:23

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