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?

Took me 2 mins and 22 seconds to solve it though this problem seemed tough to even solve at first.

stmt 2 looks the easiest..lets start with it

f - c = 3 , if f = 4 and c = 1, f - c = 3 and f + c = z = 4
if f = 5 and c = 2, f - c = 3 and f + c = z = 5, 2 values for z, not sufficient eliminate choice B and D

Stmt 1: 3a = f = 6y

now all are DISTINCT POSITIVE SINGLE DIGITS

3a = f => a = f / 3
6y = f => y = f / 6

a, y are positive digits ( 1 to 9) hence they're also single digit integers
so f is divisible by both 3 and 6
The smallest positive single digit that is divisible by 6 is 6 ITSELF

so we have f = 6, a = f/3 = 2, y = f/6 = 1

so addition is :

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

Now 1 , 2 are already used up, and c is distinct positive single digit, so is z

c cannot be greater than or equal to 4 because 4 + 6 would give 10 and we're told z is positve single digit

the only value that fits for c is 3

3 + 6 = 9 = z sufficient. answer is A
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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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How can i overlook that. Thank you so much Bunuel. Such a relief it is.
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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.
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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

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 0-9 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 f-c=3 we can get 6-c= 3 the value of z by adding 6=f and c= 3 in the units place which is z=9. Hope this helps. Cheers

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

C cant anyways be 0 since the question mentions positive numbers for all a-f,x-z

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 6
x 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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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 was unable to solve this question but then worked on it .

abc
def
xyz

Statement-1 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 non-negative 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

a-2
b
c
d
e
f-6
x
y-1
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 statement-1 is sufficient.

Statement-2

We can different values of F and C hence insufficient

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?

6. 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

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.

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.

0 is neither positive nor negative, thus neither of the digits can be zero
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This solution is not true since each of these statements is true. z = c+6 or 10+z =c +6!
However, the correct answer is A.

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