Problem 9 on FDP Question Bank

[vak3e]

First the question: How could y be 1 in this case? The only way that it could be one is if c+f > 9 ... this way the units digit could be carried over, and then in conjunction with b+e=10, you could have a y=1, maintaining the requirement that each digit is a unique value. Could someone please explain this paradox? Thank you very much! By the way, I tried to work out a conceivable solution to the problem, where every digit is in fact unique, but couldn't get it to work.

Here's the problem and solution after! Thanks for your help guys :)


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

Solution (A):

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.

[RonPurewal]

273 + 546 = 819
or
243 + 576 = 819

How do you figure b + e has to be 10? You can just as well have b + e = 11, with no carry-over from the ones place.

[vak3e]

Oh, I see! Not sure how I missed that. Thanks so much!

[RonPurewal]

Sure.

[RahulS504]

I think the answer should be:
(d) EACH statement ALONE is sufficient

This is the question below
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

This is a Data Sufficiency question; and the answer given for this is:
(a) A ALONE is sufficient but B is NOT Sufficient.

I think the answer is:
(d) EACH statement ALONE is sufficient

Explanation:
You have already proved that Statement (1) is sufficient. Below method now proves that statement (2) ALONE is also sufficient.

From (2), we have f - c = 3.
Also, from the question itself, we have c + f = z
Now when we add these two equations, we get 2f = z + 3
there is only one combination of digits that satisfy these equations, and that is : c = 3; f = 6; z = 9
Thus, statement (2) is also sufficient to get the value of z

[RonPurewal]

Also, from the question itself, we have c + f = z

No, because c + f might be more than 10.

For instance, 26 + 28 = 54. If your logic worked here, you'd be telling me that 6 + 8 = 4.

Same issue here. You can have a carry-over to the tens place. Or not.
The resulting z's will not be the same.

[RonPurewal]

Algebraically, the carry-over situation would be c + f = 10 + z, rather than just z.

(In the simple example above, 6 + 8 isn't just 4; it's 10 + 4.)

From there, you could solve in largely the same way to get a second answer.

[RahulS504]

I get it now. Thanks so Much.
:)

[RonPurewal]

You're welcome.

[LuisN65]

You're welcome.

Hi Ron,

Could you help me with this doubt?

In the explanation of the problem says that :


(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:

a b c
d e f
x y z


2 b c
d e 6
x 1 z

The question I have is how could y = 1 if all the digits are different? If y = 1 I infer that b or e must be 0 or 1 (Cero repetedor 0 and 1); but in both cases the digits are repeated... Could you help me?

Thanks a lot.

Best regards.

Luis Navarro
Looking for 700

[RonPurewal]

if you add, say, 152 and 364, what happens?