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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
 18 Jul 2010, 15:59
Question Stats:
12% (03:32) correct
87% (01:11) wrong based on 4 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 ??
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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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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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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 Posted from my mobile device 
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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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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 Very nicely explained indeed....I had picked C...
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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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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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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
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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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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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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]
07 Feb 2012, 17:08
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]
21 Dec 2012, 08:20
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: If, in the addition problem above, a, b, c, d, e, f, x, y, a
[#permalink]
21 Dec 2012, 08:20
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