## a b c + d e f ---------- x y z If, in the addition probl

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### a b c + d e f ---------- x y z If, in the addition probl

I am having trouble understanding how the equation in red below (solution section) was simplified. Your help with this would be much appreciated! (This problem is from the Manhattan Prep online question bank for Study Guide #1: Fractions, Decimals, & Percents 6th Edition)

PROBLEM:

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:
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.

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.