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If, in the addition problem above, a, b, c, d, e, f, x, y, a

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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 18 Jul 2010, 14:59
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

22% (03:15) correct 78% (01:54) wrong based on 593 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 – c = 3

How to solve this in under 2 mins ??
[Reveal] Spoiler: OA
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Re: No repeats [#permalink] New post 18 Jul 2010, 21:59
Hi,

I am posting this approach though it takes more than 2 min. Hope it is helpful to you.

From 1, we have 3a = f = 6y.
The only possible scenario is
y=1, f=6, a=2
Hence:
2bc
+ de6
------
x1z

From above, we have:
c+6=z or c+6=z+10
b+e=1(this is not possible since repetition is not allowed) or b+e=11
if c+6=z+10, then b+1+e=11
2+1+d=x

Thus the possibilities are:
c+6=z, b+e=11, 2+1+d=x (i)
or
c+6=z+10, b+e=10,2+1+d=x (ii)

From i,
c+6=z, then c=0,1,2,3.
c cannot be 0,1,2 (no repetition) hence c=3, z=9
b+e=11, the only solution is b and e are 4,7 or 7,4.
3+d=x, the only solution is 5 and 8.

From ii.
c+6=z+10
The possibilities are
c z
4 0
5 1
etc
If c=4 and z=0,
b+e=10 thus b and e are 3 and 7
3+d = x thus d and x are 5 and 8.

So we have 2 solutions for Z
z= 9 and z = 0
Thus 1 is not sufficient.


Now from 2, we have f – c = 3
f c
3 0
4 1
5 2
6 3
7 4 ... etc
3 0 cannot be the case since z will be 3 and we will have a repetition.
Considering 4 1, z will be equal to 5 and 291+384=675 is valid solution.
Considering 5 2, z will be equal to 7 and 395+412=807 is valid solution.
Hence we have 2 values of z and therefore it is not sufficient.

Taking 1 and 2 together, we have
c=3, z=9 and f=6 and therefore taking 1 and 2 together it is sufficient to answer the question.


Sorry for the mess-up and I hope this was valuable to you.

Regards,
Jack
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Re: No repeats [#permalink] New post 19 Jul 2010, 07:41
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Let me add my 2 cents...

IMO A

s1) As mentioned by jakolik
Quote:
The only possible scenario is
y=1, f=6, a=2


now c+f=z----> for z to be a positive single unit c has to be 0 or 3 ( 1 and 2 already ocuupied) and for c>3 z= two digit no. and hence violate the given condition

now if c=0 z=f which is not possible since f and z are diff

therefore c is 3 and z is 9

suff

s2) insuff
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Re: No repeats [#permalink] New post 19 Jul 2010, 17:19
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St1: 3a=f=6y
So clearly f=6,y=1,a=2
so z=9
sufficient
St2: This will produce multiple outputs for z
So Not Sufficient

Ans: A

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Re: No repeats [#permalink] New post 19 Jul 2010, 20:54
Hi guys,

I gave two solutions for the first statement 3a=f=6y, z=0 and z=9.
Hence first statement is not sufficient to answer the question.

Does any one have a better / quicker method to approach such questions?

regards,
Jack
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Re: No repeats [#permalink] New post 02 Aug 2010, 15:09
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The Reasoning is wrong because if C>3 the +1 would be transfer to the next column and z would be single unit.
The correct way is
Step 1-
From 1st statement we can fix the values of a=2,f=6 and y=1,After that the remaining numbers are 3,4,5,7,8,9.We check these values for C.

Step-2
4,5 can easily removed because Z cant be equal to 0 or 1.
If we put c=7 then z=3....... and 'b+e' should be equal to 10 which is not possible because remaining numbers are (4,5,8,9)
If we put C=8 then z=4..........remaining numbers are (3,5,7,9),b+e=10 by choosing 3 and 7,..........remaining numbers (7,9) which are not satisfying d and x.
if we put c=9 then z=5,..........remaining numbers are (3,4,7,8),b+e=10 by choosing 3 and 7,...........remaining numbers (4,8) which are not satisfying d and x.

But when we put c=3 then Z=9,remaining numbers are (4,5,7,8) and b+e=11 here so we can choose 7 and 4,remaining numbers are (5,8) which satisfies d and x.
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Re: No repeats [#permalink] New post 26 Jun 2011, 23:36
I just came across this question in the MGMAT question banks and I think that ytarun is correct in that you can't rule out c=7,8,or 9 on the basis that z is a single positive digit, which the official explanation seems to do. (If the question didn't mention "positive," then c=4, z=0 is another possibility.)

Is there a faster way of ruling out c=7,8,9 other than plugging it in and seeing if the numbers work out? It seems like a rather slow way of doing it but I can't see any other option.
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Re: No repeats [#permalink] New post 27 Jun 2011, 06:06
Let me share how we can do this in ~2 mins.

St1: 3a=f=6y => f must be a multiple of 2*3 but if we multiply 2*3 with any number other than 1, value of f will be > 9, which is not allowed. so, f=6; a=2 and y=1.

now, from the given condition c+f=z => (9-0) + 6 = z, from the set for c we can rule out values of 0, 1, 2 (as 1 and 2 are already taken and 0 will make z=6, which is not allowed as f=6) so, c can be any value from 9,8,7,5,4 => Not Sufficient

St2: f-c=3, alone this statement is Not sufficient.

St1 & St2 => c=3 => z=9 Sufficient,. Therefore answer is
[Reveal] Spoiler:
c

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Re: No repeats [#permalink] New post 27 Jun 2011, 08:52
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mn2010 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

How to solve this in under 2 mins ??


Took me >5 mins to prove
"A's" sufficiency, yet I fear that I may be missing something. Hope this is not from GPrep!!!

a b c
+
d e f
-----
x y z
-----

(1) 3a = f = 6y

y=1; f=6; a=2

2 b c
+
d e 6
-----
x 1 z
-----

Now, y=1 must be the unit's digit of 11 with 1 being carried over to the hundred's place.

So, 2+d+1=x; here "1" is the carried over 1 from y.
3+d=x
x can't be 4 as d can't be 1.
x can't be 5 as d can't be 2.
x can't be 6.
x can be 7 if d=4
x can be 8 if d=5
x can't be 9 as d can't be 6.

Let's try with x=8, d=5
2 b c
+
5 e 6
-----
8 1 z
-----

1,2,5,6,8---used
z can be 3,4,7,9
z=3, it got to be 3 of 13; So, c=7; "b+e" becomes 9+4=13 and 1 carried over from z=13; so 13+1=14; but y=1 NOT 4. Rule out.
z=4, it got to be 4 of 14; c=8(Not possible). Rule out.
z=7, c can't be 1. Rule out.
z=9, c=3; b+e=11; perfect y's "1" properly fits and 1 carries over. ------- This is first case so far.

Let's try the other scenario:

Let's try with x=7, d=4
2 b c
+
4 e 6
-----
7 1 z
-----

1,2,4,6,7 used up
3,5,8,9 left.
z=3; it must be 3 of 13; c=7(Not possible). Rule out
z=5; it must be 3 of 15; c=9. b+e+1(carried over)=8+3+1=12; but y=1. Rule out.
z=8; c=2(Not possible). Rule out.
z=9; c=3. b+e=13; but y=1. Rule out.

Thus, only one scenario fits that gave us z=9.
Sufficient.

(2) f – c = 3
a b c
+
d e f
-----
x y z
-----

f=4, c=1, z=5
2 9 1
+
3 8 4
-----
6 7 5
-----

And we already know from 1 that z=9
Not Sufficient.

Ans: "A"
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Re: No repeats [#permalink] New post 27 Jun 2011, 09:09
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I will try it in a different method... took me 2.5 mints

Problem states : All the 9 digits in the problem are different = no repeats.

(1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. if y = 2, f> 9. Since y = 1, we know that f = 6 and a = 2.

now the question reads as

2 b c
+ d e 6

= x 1 z

using our knowledge of digits we can rewrite this as

200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z
solving further ;
196 = 100(x - d) - 10(b + e) + 1(z - c)

our idea is to concentrate only on the units digit (z - c);
from the equation
6 = z - c.

lets plug in diffferent values;

z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

hence Z =9 and X = 3... perfect.


(2) INSUFFICIENT: The statement f - c = 3 .. provides us with different values... not possible.

Hence A.
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Re: No repeats [#permalink] New post 27 Jun 2011, 20:15
sudhir18n wrote:
lets plug in diffferent values;

z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

hence Z =9 and X = 3... perfect.



As I said in my previous post, I don't think you can ignore that possibility that c=7,8, or 9 which would make z= 3, 4, or 5. Thinking of that values that only fit 6 = z - c is not accurate because c + 6 can be over 10 and the 1 would carry over to the tens digit.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 07 Feb 2012, 16:08
I understood most of it apart from

lets plug in diffferent values;

z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

Any idea guys?
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 21 Dec 2012, 07:20
The key is to keep in mind that the digits are DISTINCT

c + f = z ;; Constraint => z < 9

b + e = y ;; Constraint => y < 9

a + d = x ;; Constraint => x < 9

z = ?

1) 3a = f = 6y
y cannot be greater than 1 as it would make f needs to be a single distinct digit.
y cannot be zero as then a = f = y = 0 is NOT possible as digits are distinct.

Hence
y = 1, f = 6 , a = 2

We know z<9 = c + f = c + 6 => c<3
Since digits are DISTINCT and y =1 & a = 2 are already taken values.
c = 0 or c = 3

c cannot be 0 as that would make f = z

Hence c = 3

z = 9
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Re: No repeats [#permalink] New post 26 Jun 2013, 19:43
onedayill wrote:
Let me add my 2 cents...

IMO A

s1) As mentioned by jakolik
Quote:
The only possible scenario is
y=1, f=6, a=2


now c+f=z----> for z to be a positive single unit c has to be 0 or 3 ( 1 and 2 already ocuupied) and for c>3 z= two digit no. and hence violate the given condition

now if c=0 z=f which is not possible since f and z are diff

therefore c is 3 and z is 9

suff

s2) insuff


I was stuck as hell on this question. I crossed B out because it is clearly NS. And felt that there was a trap answer in C it makes problem to easy. Using your logic I just listed out numbers 1=>9 and ticked off the numbers that have already been used. But forgot that C could not be zero which I was confused about until I reread the problem.

Since the only available numbers left are 0,3,4,5,7,8,9 I crossed out 4 through 9 because they would produce a 2 digit integer. But, through my illogical thinking I forgot that c/=0 and fell for the trap answer C in haste.

The easiest way to answer this problem is to forget about complex math, and just write down all of the digits that satisfy the fact pattern.

What I did was write out the numbers that are available, cross out the numbers that have been used, assuming a=2, f=6, y=1 and then cross out all digits that would cause a 2 digit integer.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 27 Jun 2013, 11:10
I reasoned it out, although it took me 3 min 5 seconds.

We know from A. that Y = 1, and F = 6.

There are TWO conditions: C + 6 = Z and C+6 = 10+Z.

Let's tackle the first condition:
C+6 = Z.

We can't have C=1, because we already know that Y is 1.
We can't have C = 2, because we already know that A is 2.
We can have C = 3, which would make Z = 9.

When Z = 9, that means that B + E has to equal 11, because Y = 1 (and we can't use 0, or 1 again)

Are there still ways left to have B+E = 11? We could use 7 for B, and 4 for E (or vice versa) and that would give us 11. Neither of those numbers have been used yet, so that's valid.

Now let's look at the second condition:

C + 6 = 10 + Z
We can't use C = 5, because that would make Z = 1, and 1 is already used by Y.
We can't use C = 6, because 6 is already used by F (furthermore, it would make Z = 2, and 2 is already used by A)

We can test C = 7, which would make Z = 3... but we would have to carry over the 10 (from 13) to the tens digit.
B + E + 1 (carried over from single digit 6+7=13) = 10+ 1 (which coincides with y = 1)
Which means that B+E has to equal 10.
Lets examine some of the ways that that would be possible.
We can't use 5+5, because that would use 5 twice.
We can't use 6+4, because 6 is already being used by F
We can't use 7+3, because 7 is already being used in this hypothetical test by C.
We can't use 8+2, because 2 is used by A.
We can't use 9+1, because 1 is used by Y.
We can't use 0 + 1, because we can't use 0...

Which means we can't use C=7.

The same logic applies to C=8. If C=8, Z = 4, and B+E has to equal 10
Can't use 5+5 for B+E, can't use 6+4 for B+E.
It's possible to use 7+3 for B+E, and in this case we then have to go to the hundreds digits.
If we're using 1,2,3,6,and 8 for the singles column as well as tens column, that means we would have to use 9 in the hundreds column.
This is impossible though, because if 9 is in the hundreds column then the answer would have to have four digits rather than three.
That same logic applies for the rest of the B+E answer choices if C+F (6) = 10+z

Therefore, the only possible choice left is C=3, which would make Z = 9.


A summary:
The "9" option has to be in either the tens or the single digit
C+6 < 10, otherwise, because we'd have to carry the 10 over, B+E would have to equal 10, and that's not possible without putting the 9 in the hundreds digit because if we use 9 for B or E, we'd have to use 1 for the other variable, which is impossible since 1 is already used by Y.
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Re: No repeats [#permalink] New post 31 Oct 2013, 18:50
iPinnacle wrote:
St1: 3a=f=6y
So clearly f=6,y=1,a=2
so z=9
sufficient
St2: This will produce multiple outputs for z
So Not Sufficient

Ans: A

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So whats your value for C? This seems like an unlikely method for finding a value of Z. Can you explain why Z has a value of 9?
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 26 Nov 2013, 08:21
I have a semi related question as to the digit "0" Many people are considering this as a possibility, but when I first read the question that they are all different POSITIVE single digits I excluded 0 from the options. It was my understanding that positive numbers are > 0 and that 0 is neither + or -.

Can someone shed light on this?
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 26 Nov 2013, 09:14
Expert's post
evangreen wrote:
I have a semi related question as to the digit "0" Many people are considering this as a possibility, but when I first read the question that they are all different POSITIVE single digits I excluded 0 from the options. It was my understanding that positive numbers are > 0 and that 0 is neither + or -.

Can someone shed light on this?


Yes, 0 is neither positive nor negative, thus neither of the digits can be zero.

Similar questions to practice:
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if-x-0-rstu-where-r-s-t-and-u-each-represent-a-nonzero-109735.html
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if-y-0-jkmn-where-j-k-m-and-n-each-represent-a-nonzero-130018.html
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if-a-0-abc-where-a-b-and-c-are-digits-of-a-is-a-greater-138282.html
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Hope this helps.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 08 Dec 2013, 04:25
What is the final verdict on this one? Its A or C?

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: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink] New post 08 Dec 2013, 05:23
Expert's post
sumitaries wrote:
What is the final verdict on this one? Its A or C?

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


The OA is given under the spoiler in the first post. It's A.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a   [#permalink] 08 Dec 2013, 05:23
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