A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?

• Let the 2-digit number be $10a + b$, where $a$ and $b$ are digits and $a > b$ to ensure the original number is larger.

• The reversed number is $10b + a$. The division is $(10a + b) \div (10b + a)$.

• To maximize the remainder, the divisor ($10b + a$) should be as large as possible while the quotient remains 1 (since a quotient of 2 or more would require the larger number to be at least double the smaller, which is impossible for 2-digit reversals unless the smaller is very small, limiting the remainder).

• The maximum possible divisor occurs when $b=9$ and $a=8$ (since $a > b$), giving the numbers 89 and 98.

• Dividing 98 by 89 gives a quotient of 1 and a remainder of $98 - 89 = 9$. However, we must check if a larger remainder is possible with a smaller divisor.

• Consider the case where the larger number is 98 and the smaller is 89? No, 98 > 89. Let's test the maximum difference. The remainder $R = (10a+b) - (10b+a) = 9(a-b)$ if the quotient is 1. The maximum value of $a-b$ is 9 (e.g., $a=9, b=0$). If $a=9, b=0$, numbers are 90 and 09 (9). $90 \div 9 = 10$ remainder 0. This logic holds only if quotient is 1.

• Let's re-evaluate. We want to maximize $R$ in $L = Q \times S + R$. Since $L < 2S$ is not always true, let's look at specific cases. If $S$ is small, $R$ can be large. Max $S$ is 98? No, $S$ is the smaller. Min $S$ is 10? No, min 2-digit is 10. If $S=10, L=01$ (invalid). If $S=12, L=21$. $21 \div 12 = 1$ rem 9. If $S=19, L=91$. $91 \div 19 = 4$ rem 15. If $S=29, L=92$. $92 \div 29 = 3$ rem 5. If $S=39, L=93$. $93 \div 39 = 2$ rem 15. If $S=49, L=94$. $94 \div 49 = 1$ rem 45. Here $L=94, S=49$. $94 = 1 \times 49 + 45$. The remainder is 45. Can it be higher? Max remainder is $S-1$. If $S=49$, max rem is 48. We got 45. If $S=59, L=95$. $95 \div 59 = 1$ rem 36. If $S=69, L=96$. $96 \div 69 = 1$ rem 27. If $S=79, L=97$. $97 \div 79 = 1$ rem 18. If $S=89, L=98$. $98 \div 89 = 1$ rem 9. The largest remainder found is 45.