**The sum of the digits of a number N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?**

Let's analyze the problem step-by-step to find the remainder when \( N \) is divided by 33, given the following conditions:

1. **The sum of the digits of \( N \) is 23.**
2. **The remainder when \( N \) is divided by 11 is 7.**

### Step 1: Understand the Problem
- We know that when a number is divided by 33, it can be expressed as a combination of its remainders when divided by 11 and 3 (since \(33 = 3 \times 11\)). This is because 3 and 11 are coprime.

### Step 2: Use the Information About Modulo 11

Given:
\[
N \equiv 7 \pmod{11}
\]

This tells us that when \( N \) is divided by 11, the remainder is 7.

### Step 3: Determine \( N \mod 3 \)

The sum of the digits of \( N \) is 23. A key property of modulo 3 is that a number and the sum of its digits have the same remainder when divided by 3. Therefore:
\[
N \equiv 23 \pmod{3}
\]

Calculating \( 23 \mod 3 \):
\[
23 \div 3 = 7 \text{ remainder } 2
\]
So:
\[
N \equiv 2 \pmod{3}
\]

### Step 4: Apply the Chinese Remainder Theorem

Now we have two congruences:
\[
N \equiv 7 \pmod{11}
\]
\[
N \equiv 2 \pmod{3}
\]

We want to find \( N \pmod{33} \). Let's denote this remainder by \( r \).

### Step 5: Solve for \( N \pmod{33} \)

To find \( r \), we need to solve the system of congruences:
1. \( r \equiv 7 \pmod{11} \)
2. \( r \equiv 2 \pmod{3} \)

Consider numbers of the form \( r = 11k + 7 \) (since \( r \equiv 7 \pmod{11} \)). We need this to satisfy the second congruence:
\[
11k + 7 \equiv 2 \pmod{3}
\]

Calculating \( 11 \mod 3 \):
\[
11 \equiv 2 \pmod{3}
\]

Substitute this into the equation:
\[
2k + 7 \equiv 2 \pmod{3}
\]

Subtract 7 from both sides:
\[
2k \equiv -5 \pmod{3}
\]

Since \( -5 \equiv 1 \pmod{3} \), we have:
\[
2k \equiv 1 \pmod{3}
\]

To solve for \( k \), we need the multiplicative inverse of 2 modulo 3, which is 2 (since \( 2 \times 2 = 4 \equiv 1 \pmod{3} \)). Thus:
\[
k \equiv 2 \pmod{3}
\]

This implies \( k = 3m + 2 \) for some integer \( m \). Substitute back for \( r \):
\[
r = 11(3m + 2) + 7 = 33m + 22 + 7 = 33m + 29
\]

Thus:
\[
r \equiv 29 \pmod{33}
\]

### Conclusion

The remainder when \( N \) is divided by 33 is \(\boxed{29}\).

Would you like further details or have any questions?

### Related Questions:
1. How do you find the multiplicative inverse of a number modulo another?
2. What is the Chinese Remainder Theorem and how is it applied?
3. Can a number have different remainders when divided by different numbers? How do you find them?
4. How can properties of modular arithmetic be used in solving cryptographic problems?
5. What are some practical applications of the Chinese Remainder Theorem?

### Tip:
When working with modular arithmetic, breaking down a problem into smaller congruences often simplifies finding a solution.

Find the remainder when a number N, with sum of digits 23 and remainder 7 when divided by 11, is divided by 33 using modular arithmetic and the Chinese Remainder Theorem.

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