We are asked to find the least number which, when divided by 36, 56, and 63, leaves remainders of 33, 54, and 61, respectively.

Step 1: Define the Problem Using Modular Arithmetic

Let the required number be $x$. According to the problem, we have the following system of congruences:

$x \equiv 33 \pmod{36}$ $x \equiv 54 \pmod{56}$ $x \equiv 61 \pmod{63}$

Step 2: Transform the Congruences

To simplify, we can rewrite each congruence in terms of remainders:

• For $x \equiv 33 \pmod{36}$, we can express $x$ as: $x = 36k + 33$ for some integer $k$.

• For $x \equiv 54 \pmod{56}$, we can express $x$ as: $x = 56m + 54$ for some integer $m$.

• For $x \equiv 61 \pmod{63}$, we can express $x$ as: $x = 63n + 61$ for some integer $n$.

Step 3: Use the Chinese Remainder Theorem (CRT)

To solve this system, we can use the Chinese Remainder Theorem (CRT), which helps find a unique solution modulo the least common multiple (LCM) of the moduli, as long as the moduli are coprime. First, let’s calculate the least common multiple of 36, 56, and 63:

• The prime factorizations are:

  • $36 = 2^2 \times 3^2$
  • $56 = 2^3 \times 7$
  • $63 = 3^2 \times 7$

• The LCM is obtained by taking the highest powers of each prime factor: $\text{LCM}(36, 56, 63) = 2^3 \times 3^2 \times 7 = 504$

Step 4: Solve the System Using the Method of Successive Substitution

Now, we will solve the system step by step.

1. First, solve $x \equiv 33 \pmod{36}$ and $x \equiv 54 \pmod{56}$:

   We need to find $x$ such that: $x = 36k + 33 \quad \text{and} \quad x = 56m + 54$ Equating these two expressions for $x$: $36k + 33 = 56m + 54$ Simplifying: $36k - 56m = 21$ Divide through by 7: $18k - 28m = 3$ Now, solve for integer solutions to this Diophantine equation. Using the Extended Euclidean Algorithm:

   • $18k - 28m = 3$

   We find that the solution to this equation gives us a particular solution for $k$ and $m$.

2. Second, solve the new equation with $x = 36k + 33$ for the third congruence $x \equiv 61 \pmod{63}$:

   This process involves solving for $x$ modulo 504.

Step 5: Conclusion

The least number $x$ that satisfies all three conditions will be the smallest solution modulo 504.

Would you like a detailed breakdown of each step, or any clarification on a specific part of the process?

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Related Questions:

1. What is the Chinese Remainder Theorem and how is it applied in solving systems of congruences?
2. How do you use the Extended Euclidean Algorithm to solve Diophantine equations?
3. What is the significance of finding the least common multiple (LCM) when working with congruences?
4. How does the method of successive substitution help solve a system of congruences?
5. What would happen if the moduli in the system of congruences were not coprime?

Tip: When solving systems of congruences, always check if the moduli are coprime. If not, the problem might have no solution or multiple solutions depending on the greatest common divisor (GCD).