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

The correct option is 45.

Step-by-step Explanation:

Let the 2-digit number be represented as $10a+b$, where $a$ and $b$ are digits ranging from 1 to 9 (since reversing the number also creates a valid 2-digit number, neither digit can be 0).

When the number is reversed, the new number is $10b+a$.

We are looking for the largest possible remainder when the larger of the two numbers is divided by the smaller one.

For any division operation, the remainder $R$ must strictly be less than the divisor $D$ (the smaller number). That is, $R<D$. To maximize the remainder $R$, we need a relatively large divisor where the quotient is 1, so that the remainder is simply $\text{Larger Number}-\text{Smaller Number}$.

Let's calculate the difference between the larger number and the smaller number:

$\text{Difference}=(10a+b)-(10b+a)=9(a-b)$

Since $a$ and $b$ are single digits (1 to 9), the maximum difference between two digits $a-b$ can be $9-1=8$, giving a maximum difference of $9\times 8=72$ (for numbers 91 and 19). However, dividing 91 by 19 yields $91=19\times 4+15$, which gives a remainder of 15.

Let's test pairs of numbers with smaller quotients to find the largest remainder:

1. Consider $a-b=5$:

Take the pair 94 and 49:

$94=49\times 1+45$

Here, the remainder is 45, which is less than the divisor 49, so it is a valid remainder.

2. Let's check if any remainder larger than 45 is possible:

If quotient = 1, then $\text{Remainder}=\text{Larger}-\text{Smaller}=9(a-b)$.

For remainder to be greater than 45, $9(a-b)$ must be at least 54 (meaning $a-b\ge 6$):

- For $a-b=6$: largest pair is 93 and 39. Difference is $54$. Dividing 93 by 39 gives quotient 2 and remainder $93-2\times 39=15$.

- For $a-b=7$: largest pair is 92 and 29. Difference is $63$. Dividing 92 by 29 gives quotient 3 and remainder $92-3\times 29=5$.

- For $a-b=8$: largest pair is 91 and 19. Difference is $72$. Dividing 91 by 19 gives quotient 4 and remainder $91-4\times 19=15$.

Thus, the maximum achievable remainder is 45 (obtained from dividing 94 by 49).