Let's go through the simplification of the given sum again step-by-step:

1. Understand the Problem:
   The problem involves determining the remainder when a number $N$ is divided by 33, given that:

   • The sum of its digits is 23.
   • The remainder when $N$ is divided by 11 is 7.

2. Key Observations Using Modulo Properties:

   • Since the remainder when $N$ is divided by 11 is 7: $N \equiv 7 \pmod{11}$
   • Since the sum of the digits of $N$ is 23, and a number and the sum of its digits are congruent modulo 3: $N \equiv 23 \pmod{3}$ Simplifying $23 \mod 3$: $23 = 3 \times 7 + 2 \implies N \equiv 2 \pmod{3}$

3. Apply the Chinese Remainder Theorem (CRT):

   • We have the system of congruences: N \equiv 7 \pmod{11} \\ N \equiv 2 \pmod{3} \end{cases}$$
   • We seek $N \equiv x \pmod{33}$, where 33 is the product of 3 and 11 (since 3 and 11 are coprime).

4. Solve for $N \pmod{33}$:

   • Start by expressing $N$ in terms of modulo 11: $N = 11k + 7$ Now, substitute this into the modulo 3 congruence: $11k + 7 \equiv 2 \pmod{3}$

   • Since $11 \equiv 2 \pmod{3}$, this reduces to: $2k + 7 \equiv 2 \pmod{3}$

   • Simplify by subtracting 7: $2k \equiv -5 \pmod{3}$

   • Since $-5 \equiv 1 \pmod{3}$: $2k \equiv 1 \pmod{3}$

   • The multiplicative inverse of 2 modulo 3 is 2 (because $2 \times 2 = 4 \equiv 1 \pmod{3}$): $k \equiv 2 \pmod{3}$

   • Write $k$ in terms of modulo 3: $k = 3m + 2$

   • Substitute back into $N = 11k + 7$: $N = 11(3m + 2) + 7 = 33m + 22 + 7 = 33m + 29$

   • Therefore, $N \equiv 29 \pmod{33}$.

5. Conclusion:
   The remainder when $N$ is divided by 33 is $\boxed{29}$.

Feel free to ask for any clarifications or more details!

Related Questions:

1. How does the Chinese Remainder Theorem work, and when is it applicable?
2. What is the significance of modular inverses in number theory?
3. How can one solve more complex systems of congruences with more than two conditions?
4. What are other methods to simplify sums or solve congruences?
5. How is modular arithmetic used in cryptography?

Tip:

When faced with modular arithmetic problems, always look for patterns and properties like those of remainders and use the Chinese Remainder Theorem to simplify complex problems step-by-step.