What is the largest two-digit number that gives 10 as the remainder when divided by 11?

To solve this, we are looking for the largest two-digit number, \( N \), that gives a remainder of 10 when divided by 11. Mathematically, this is expressed as:

\[
N \equiv 10 \ (\text{mod} \ 11)
\]

This means that \( N \) can be written in the form:

\[
N = 11k + 10
\]

where \( k \) is an integer.

Now, since \( N \) is a two-digit number, we know:

\[
10 \leq N \leq 99
\]

We now find the largest possible \( k \) such that \( N \) remains a two-digit number. To do this:

\[
11k + 10 \leq 99
\]

First, subtract 10 from both sides:

\[
11k \leq 89
\]

Now, divide by 11:

\[
k \leq \frac{89}{11} \approx 8.09
\]

Thus, the largest integer \( k \) is 8. Plugging this value of \( k \) back into the equation for \( N \):

\[
N = 11(8) + 10 = 88 + 10 = 98
\]

Therefore, the largest two-digit number that gives a remainder of 10 when divided by 11 is **98**.

Would you like more details or have any questions?

Here are 5 related questions to expand your knowledge:
1. What is the smallest two-digit number that gives a remainder of 10 when divided by 11?
2. How would you solve this if you were looking for a remainder of 5 when divided by 11?
3. Can you generalize a method to find a number with a specific remainder when divided by any integer?
4. What are some practical applications of modular arithmetic?
5. What is the largest three-digit number that gives 10 as a remainder when divided by 11?

**Tip**: When dealing with modular arithmetic, always check both upper and lower bounds to ensure the number falls within the desired range.

Find the largest two-digit number that gives a remainder of 10 when divided by 11 using modular arithmetic. This solution involves key concepts in number theory and modular arithmetic, suitable for students in Grades 8-10.

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