Let the original number be $10a + b$, where $a$ is the tens digit and $b$ is the ones digit. Reversed number = $10b + a$. Assume $10a + b > 10b + a$, so the larger number is $10a + b$.

The remainder when dividing the larger number by the smaller one is:

$(10a + b) \div (10b + a)$

To maximize the remainder, set $a = 9$ (largest possible tens digit). Original number = $90 + b$ Reversed number = $10b + 9$

Now, test different values of $b$:

1. If $b = 4$:

• Original number = $94$
• Reversed number = $49$
• Division: $94 \div 49$ → Quotient = $1$, Remainder = $94 - 49 = 45$

2. If $b = 5$:

• Original number = $95$
• Reversed number = $59$
• Division: $95 \div 59$ → Quotient = $1$, Remainder = $95 - 59 = 36$

3. If $b = 6$:

• Original number = $96$
• Reversed number = $69$
• Division: $96 \div 69$ → Quotient = $1$, Remainder = $96 - 69 = 27$

4. If $b = 7$:

• Original number = $97$
• Reversed number = $79$
• Division: $97 \div 79$ → Quotient = $1$, Remainder = $97 - 79 = 18$

5. If $b = 8$:

• Original number = $98$
• Reversed number = $89$
• Division: $98 \div 89$ → Quotient = $1$, Remainder = $98 - 89 = 9$

The largest remainder is $45$, which occurs when the original number is $94$ and the reversed number is $49$.

Thus, the largest possible remainder is $45$.