In standard arithmetic, $ab$ means $a \times b$. But when $ab$ represents a 2-digit number with tens digit $a$ and ones digit $b$ (like $54$), it actually means:
$$ab = 10 \times a + b = 10a + b$$
If you reverse the digits to get $ba$ (like $45$), its value is: $$ba = 10 \times b + a = 10b + a$$
Theorem 1: The Difference of Reversible Numbers is Always Divisible by 9
$$(10a + b) - (10b + a) = 10a - a + b - 10b = 9a - 9b = \mathbf{9(a - b)}$$
Because $9$ is a factor, the difference is ALWAYS divisible by $9$. Dividing by $9$ leaves $(a - b)$, the difference between the two digits!
Theorem 2: The Sum of Reversible Numbers is Always Divisible by 11
$$(10a + b) + (10b + a) = 10a + a + b + 10b = 11a + 11b = \mathbf{11(a + b)}$$
Because $11$ is a factor, the sum is ALWAYS divisible by $11$. Dividing by $11$ leaves $(a + b)$, the sum of the two digits!
Example: Take the number $83$. Add it to its reverse, and divide the sum by $11$. What is the quotient?
Step 1: Reverse of $83$ is $38$.
Step 2: Sum $= 83 + 38 = 121$.
Step 3: Divide by $11$: $\frac{121}{11} = \mathbf{11}$. Notice this equals the sum of the digits: $8 + 3 = 11$!
In proving why reversing a 3-digit number $abc$ ($100a + 10b + c$) works, students often write $abc - cba$ as $0$.
Reality: $abc - cba = (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c)$, which is always a multiple of $99$!
Bank accounting systems use the "divisible by 9" rule to catch transposition errors—when an accountant accidentally enters $54$ instead of $45$, the difference is divisible by $9$, alerting the fraud detection software instantly.