Follow the steps below.
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 \(F\).
What is the connection between the number \(F\) and the numbers \(A\) and \(B\)?
Try the same steps with different numbers. Do you think you will always get the same result? Why or why not?
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 \(A-B=F\).
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, \(F=A-B\).