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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Sep 2014, 21:10
Bunuel wrote: 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. Hi Bunuel. My understanding is: From the first statement we have a=2 f=6 y=1 Now with hit and trial, we have following two possible sets for rest of the integers 1. c=3 z=9 b/e=7/4 d=5 x=8 273 243 +546  819 or 243 +576  819 2. c=9 z=4 b/e=7/3 d=5 x=8 279 +536  815 or 239 +576  815 Please tell me how option A is sufficient.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Sep 2014, 21:19
aadikamagic wrote: Bunuel wrote: 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. Hi Bunuel. My understanding is: From the first statement we have a=2 f=6 y=1 Now with hit and trial, we have following two possible sets for rest of the integers 1. c=3 z=9 b/e=7/4 d=5 x=8 273 243 +546  819 or 243 +576  819 2. c=9 z=4 b/e=7/3 d=5 x=8 279 + 536  81 5 or 239 + 576  81 5Please tell me how option A is sufficient. We are given that a, b, c, d, e, f, x, y, and z each represent different positive single digits
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Sep 2014, 23:39
How can i overlook that. Thank you so much Bunuel. Such a relief it is.
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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Oct 2014, 06:15
I think both together are sufficient. Z=9. Can you tell me why z should be something else. The numbers should be different. Where am I wrong ?



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Oct 2014, 07:21



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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05 May 2015, 02:44
1) 3a = f = 6y so, f=6, a=2 and y=1 now, c+f=z , or (c+6=z)  z can take values 6, 7, 8, 9
now, 6 is taken by f z cant be 7 becoz c cant be 1 (since y=1) z cant be 8 becoz c cant be 2 (since a=2) only value left for z = 9  (1) is sufficient



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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05 May 2015, 03:00
As the first statement says 3a=f=6y it means f is divisible by both 3 and 6. Since it is divisible by 6 only number between 09 divisible by 6 is 6 itself. That gives us the value for f. Since no other information can be found from st1 its not sufficient on its own.
Moving to st2 it doesn't say much apart from the difference which can be a combination of any possible numbers having a difference of 3. Not sufficient.
Combining both statements fc=3 we can get 6c= 3 the value of z by adding 6=f and c= 3 in the units place which is z=9. Hope this helps. Cheers



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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06 Dec 2015, 23:58
Since each alphabet represents a single digit, 6 times a digit clearly implies that the maximum value it can have is 6 [because 6 x 2 = 12] That means we get y =1, f = 6 and a = 3
So the problem now reads
3bc + ae6  x1z  If c =2, then z = 8 Since addition of b and e yields 1, it means that actual addition yields 11 1,2,3,6,8 are taken so only 4,5,7 and 9 remain 7 and 4 add up to 11 so we take b = 7 and e = 4 [our concern is the one's place of the final addition AND ensure that each alphabet has a unique value]
So the problem now reads 372 + a46  x18  a cannot be 9 as the addition completes in a 3 digit number Only 5 is the digit that remains that will satisfy the conditions Thus we can solve the problem with the 1st statement
The 2nd statement only gives us the difference between the one's places and throws up multiple options
Hence A



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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15 Dec 2015, 20:26
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 C cant anyways be 0 since the question mentions positive numbers for all af,xz



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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27 Jan 2017, 08:40
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 ?? The problem states that all 9 single digits in the problem are different; in other words, there are no repeated digits. (1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. Any greater value for y, such as y = 2, would make f greater than 9. Since y = 1, we know that f = 6 and a = 2. We can now rewrite the problem as follows: 2 b c + d e 6x 1 z In order to determine the possible values for z in this scenario, we need to rewrite the problem using place values as follows: 200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z This can be simplified as follows: 196 = 100(x – d) – 10(b + e) + 1(z – c) Since our focus is on the units digit, notice that the units digit on the left side of the equation is 6 and the units digit on the right side of the equation is (z – c). Thus, we know that 6 = z – c. Since z and c are single positive digits, let's list the possible solutions to this equation. z = 9 and c = 3 z = 8 and c = 2 z = 7 and c = 1 However, the second and third solutions are NOT possible because the problem states that each digit in the problem is different. The second solution can be eliminated because c cannot be 2 (since a is already 2). The third solution can be eliminated because c cannot be 1 (since y is already 1). Thus, the only possible solution is the first one, and so z must equal 9. (2) INSUFFICIENT: The statement f – c = 3 yields possible values of z. For example f might be 7 and c might be 4. This would mean that z = 1. Alternatively, f might be 6 and c might be 3. This would mean that z = 9. The correct answer is A.
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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28 Feb 2017, 14:04
Honestly this seems to be one of those time drainer questions. Just admit that an average person doesn't have the brainpower for this, guess and move on. Its fine to miss something like this.



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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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28 Feb 2017, 14:26
iliavko wrote: Honestly this seems to be one of those time drainer questions. Just admit that an average person doesn't have the brainpower for this, guess and move on. Its fine to miss something like this. I concur. This is an EXCELLENT candidate for guessing and moving on, ESPECIALLY if you are at all behind in time. Keep in mind that you can answer several questions incorrectly and still get a 99th percentile score. Cheers, Brent
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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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21 Jul 2017, 01:25
I was unable to solve this question but then worked on it .
abc def xyz
Statement1 3a=f=6y
a, f and y can be any numbers out of 1 to 9 which match the given condition zero is not included since the question stem says +ve integers. (not nonnegative integers in which case we will have to take zero as well). y=f/6 and a=f/3. f is a distinct integer hence it has to be divisible by both 6 and 3. we can plugin and check and the no is 6. If f=6 than a=6/3=2 and y=1.
Plugin the values
2bc de6 x1z
Write down the nos against given variables a2 b c d e f6 x y1 z
Since nos are distinct nos available are 4, 3, 5, 7, 8,9
2bc de6 x1z
Z cannot be 0 (mentioned in the question stem).
This means that C can be only 3 since 1, 2 are already taken and it cannot be a digit which is greater than 3 since the sum will lead to value of Z which is not allowed or already taken.
Hence statement1 is sufficient.
Statement2
We can different values of F and C hence insufficient



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If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]
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04 Aug 2017, 08:52
Bunuel wrote: Babken37 wrote: I think both together are sufficient. Z=9. Can you tell me why z should be something else. The numbers should be different. Where am I wrong ? The statement f  c = 3 yields different values of z. For example f might be 7 and c might be 4. This would mean that z = 1. Alternatively, f might be 6 and c might be 3. This would mean that z = 9. Bunuel, could you explain to me why C+6 cannot equal 10(S) + Z. I understand that the problem says that each digit has to a be positive single digit, however: 1) the way that I interpret that, that just means that the unit digits of C+6 is less than 10, and that there may be carry over to the next column. 2) However, if we are to accept that positive single digit means that C+6=<9, then the next column makes no sense because there is no way for B+e<=1 when neither B or E can equal 1. B or E must equal 10(q) + 1 (1 being the units digit), thus if it this is true, then this means the wording is inconsistent. Am I missing something obvious here?




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