# Problem of the Week Problem A and Solution Adding and Subtracting

## Problem

Step 1: Pick two different three-digit numbers. Label the larger number $$A$$ and the smaller number $$B$$.

Step 2: Subtract $$B$$ from $$999$$ and label the difference $$C$$.

Step 3: Add $$A$$ and $$C$$ and label the sum $$D$$.

Step 4: Subtract $$1000$$ from $$D$$ and label the difference $$E$$.

Step 5: Add $$1$$ to $$E$$ and label the sum $$F$$.

1. What is the connection between the number $$F$$ and the numbers $$A$$ and $$B$$?

2. Try the same steps with different numbers. Do you think you will always get the same result? Why or why not?

## Solution

1. We can work through this procedure with any two, random, three-digit numbers. Let’s try starting with $$814$$ and $$275$$.

Step 1: The larger number $$814$$ is labelled $$A$$, and the smaller number $$275$$ is labelled $$B$$.

Step 2: The difference $$999-B$$ is $$999 - 275 = 724$$, and is labelled $$C$$.

Step 3: The sum $$A+C$$ is $$814 + 724 = 1538$$, and is labelled $$D$$.

Step 4: The difference $$D-1000$$ is $$1538-1000 = 538$$, and is labelled $$E$$.

Step 5: The sum $$E+1$$ is $$538+1 = 539$$, and is labelled $$F$$.

Notice that $$814 - 275 = 539$$. So $$A-B=F$$.

2. If you try this procedure with any two three-digit numbers, it will always work out that $$A-B=F$$. We will show this using algebra.

Step 1: We choose three-digit numbers $$A$$ and $$B$$ so that $$A\geq B$$.

Step 2: We calculate $$999-B$$, and label this $$C$$. That is, $$C=999-B$$.

Step 3: We add $$A+C$$. So, we calculate $$A+999-B$$, and label this $$D$$. That is, $$D = A + 999 - B$$.

Step 4: We subtract $$1000$$ from $$D$$. So, we calculate $$A + 999 - B -1000$$, and label this $$E$$. That is, $$E = A + 999 - B -1000$$.

Step 5: We add $$1$$ to $$E$$. So, we calculate $$A + 999 - B -1000 + 1$$, and label this $$F$$. That is, $$F = A + 999 - B -1000 + 1$$.

Now, we can simplify this expression to get $$A-B+999-1000+1=A-B$$, since $$999-1000+1=0$$. So, $$F=A-B$$.