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Magazine Chapter 4: Problem 20
Use the construction in the proof of the Chinese remainder theorem to find all solutions to the system of congruences \(x \equiv 2(\bmod 3), x \equiv 1(\bmod 4),\) and \(x \equiv 3(\bmod 5)\)
Short Answer
Expert verified
The solution to the system is \(x \equiv 53 (\bmod 60)\).
Step by step solution
01
Understanding the System of Congruences
Given the system of congruences: 1. \(x \equiv 2 \bmod 3\) 2. \(x \equiv 1 \bmod 4\) 3. \(x \equiv 3 \bmod 5\)
02
Verify Pairwise Coprimality of Moduli
Check that the moduli (3, 4, and 5) are pairwise coprime. This is true because: \(\gcd(3, 4) = 1, \gcd(4, 5) = 1,\) and \(\gcd(3, 5) = 1\).
03
Compute Product of Moduli
Calculate the product of the moduli: \(M = 3 \times 4 \times 5 = 60\)
04
Calculate Partial Products
Compute the partial products: \(M_1 = \frac{M}{3} = \frac{60}{3} = 20\) \(M_2 = \frac{M}{4} = \frac{60}{4} = 15\) \(M_3 = \frac{M}{5} = \frac{60}{5} = 12\)
05
Compute Inverses of Partial Products
Find the inverses of the partial products modulo each respective modulus: 1. Find the inverse of 20 modulo 3:\(20 \bmod 3 = 2\). Solve \(2y \equiv 1 (\bmod 3)\) to get \(y=2\). Thus, \(M_1^{-1} = 2\).2. Find the inverse of 15 modulo 4:\(15 \bmod 4 = 3\). Solve \(3y \equiv 1 (\bmod 4)\) to get \(y=3\). Thus, \(M_2^{-1} = 3\).3. Find the inverse of 12 modulo 5:\(12 \bmod 5 = 2\). Solve \(2y \equiv 1 (\bmod 5)\) to get \(y=3\). Thus, \(M_3^{-1} = 3\).
06
Calculate the Solution Using CRT Formula
Use the Chinese Remainder Theorem formula to calculate the solution: \(x \equiv a_1 M_1 M_1^{-1} + a_2 M_2 M_2^{-1} + a_3 M_3 M_3^{-1} (\bmod M)\), where \(a_1 = 2\), \(a_2 = 1\), \(a_3 = 3\). Thus, \(x \equiv 2 \times 20 \times 2 + 1 \times 15 \times 3 + 3 \times 12 \times 3 \bmod 60\) Calculate each term: \(2 \times 20 \times 2 = 80\) \(1 \times 15 \times 3 = 45\) \(3 \times 12 \times 3 = 108\) Sum them up and take modulo 60: \(80 + 45 + 108 = 233\) \(233 \bmod 60 = 53\)
07
Write General Solution
The solution to the system of congruences is given by: \(x \equiv 53 (\bmod 60)\) Thus, the general solution is \(x = 53 + 60k\), for any integer \(k\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Congruences
A system of congruences involves multiple congruence equations that need to be solved simultaneously. In the given exercise, we have three congruences:
1. \(x \equiv 2 (\bmod 3)\)
2. \(x \equiv 1 (\bmod 4)\)
3. \(x \equiv 3 (\bmod 5)\)
To find a solution that satisfies all three congruences, we need to find a number \(x\) that fits all these modular requirements. This problem is a classic example of using the Chinese Remainder Theorem (CRT) to find a unique solution modulo the product of the moduli.
1. \(x \equiv 2 (\bmod 3)\)
2. \(x \equiv 1 (\bmod 4)\)
3. \(x \equiv 3 (\bmod 5)\)
To find a solution that satisfies all three congruences, we need to find a number \(x\) that fits all these modular requirements. This problem is a classic example of using the Chinese Remainder Theorem (CRT) to find a unique solution modulo the product of the moduli.
Pairwise Coprime Moduli
For the Chinese Remainder Theorem to work, the moduli in the system of congruences must be pairwise coprime. This means that the greatest common divisor (GCD) of any two moduli must be 1.
Let's verify this for our moduli 3, 4, and 5:
Since all pairs have a GCD of 1, the moduli are pairwise coprime. This ensures that a unique solution exists for the system modulo the product of the moduli, which in our case is \(3 \times 4 \times 5 = 60\).
Let's verify this for our moduli 3, 4, and 5:
- \(\gcd(3, 4) = 1\)
- \(\gcd(4, 5) = 1\)
- \(\gcd(3, 5) = 1\)
Since all pairs have a GCD of 1, the moduli are pairwise coprime. This ensures that a unique solution exists for the system modulo the product of the moduli, which in our case is \(3 \times 4 \times 5 = 60\).
Partial Products Inversion
To solve the system using the Chinese Remainder Theorem, we need to compute partial products and their inverses. Here's how it's done:
Using these inverses, we can form the solution using the formula:
\(x \equiv a_1 M_1 M_1^{-1} + a_2 M_2 M_2^{-1} + a_3 M_3 M_3^{-1} (\bmod M)\)
Substituting the values from our exercise, we get:
\(x \equiv 2 \times 20 \times 2 + 1 \times 15 \times 3 + 3 \times 12 \times 3 \bmod 60\)
Calculate the components:
\(x \equiv 53 (\bmod 60)\)
Or in general form:
\(x = 53 + 60k\) for any integer \(k\).
- First, compute the total product of the moduli: \(M = 60\).
- Next, compute the partial products for each congruence:
\(M_1 = \frac{M}{3} = 20\)
\(M_2 = \frac{M}{4} = 15\)
\(M_3 = \frac{M}{5} = 12\) - Then, find the inverse of each partial product modulo the corresponding modulus:
Inverse of 20 modulo 3: Solve \(20 \bmod 3 = 2\), we need \(2y \equiv 1 (\bmod 3)\). The solution is \(y = 2\), so \(M_1^{-1} = 2\).
Inverse of 15 modulo 4: Solve \(15 \bmod 4 = 3\), we need \(3y \equiv 1 (\bmod 4)\). The solution is \(y = 3\), so \(M_2^{-1} = 3\).
Inverse of 12 modulo 5: Solve \(12 \bmod 5 = 2\), we need \(2y \equiv 1 (\bmod 5)\). The solution is \(y = 3\), so \(M_3^{-1} = 3\).
Using these inverses, we can form the solution using the formula:
\(x \equiv a_1 M_1 M_1^{-1} + a_2 M_2 M_2^{-1} + a_3 M_3 M_3^{-1} (\bmod M)\)
Substituting the values from our exercise, we get:
\(x \equiv 2 \times 20 \times 2 + 1 \times 15 \times 3 + 3 \times 12 \times 3 \bmod 60\)
Calculate the components:
- \(2 \times 20 \times 2 = 80\)
- \(1 \times 15 \times 3 = 45\)
- \(3 \times 12 \times 3 = 108\)
\(x \equiv 53 (\bmod 60)\)
Or in general form:
\(x = 53 + 60k\) for any integer \(k\).
