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Find a simultaneous solution for the system of four congruences: $$ \begin{aligned} x & \equiv 1(\bmod 2) \\ x & \equiv 2(\bmod 3) \\ x & \equiv 3(\bmod 5) \\ x & \equiv 5(\bmod 7) \end{aligned} $$

Short Answer

Expert verified
The simultaneous solution for the system of congruences is \(x = 53\) for any modulus \(210\).

Step by step solution

01

Solving Individual Congruences

Start by finding an integer solution \(x\) to each congruence individually. In this case, each congruence is simple. Therefore, for \(x \equiv 1 \mod 2\), any odd number will suffice. For \(x \equiv 2 \mod 3\), you can choose \(2\). For \(x \equiv 3 \mod 5\), \(3\) is an apt choice. Finally, for \(x \equiv 5 \mod 7\), you can take \(5\).
02

Construct combined congruences

Next, each individual solution needs to be combined to form a system of two congruences. That is, the first two congruences \(x \equiv 1 \mod 2\) and \(x \equiv 2 \mod 3\) can be combined to get \(x \equiv 5 \mod 6\) (since \(2*3=6\)). Similarly, the next two congruences \(x \equiv 3 \mod 5\) and \(x \equiv 5 \mod 7\) can be combined to form \(x \equiv 8 \mod 35\) (since \(5*7=35\)).
03

Final Solution

Finally, the combined congruences obtained in Step 2 can again be combined. We have two congruences \(x \equiv 5 \mod 6\) and \(x \equiv 8 \mod 35\). We will combine these to find the simultaneous solution for the initial four congruences. The least common multiple of modulus 6 and 35 is 210. Therefore, our final solution is \(x \equiv 53 \mod 210\). Hence, the solution for the problem is \(x = 53\) for any modulus \(210\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Congruence Arithmetic
Congruence arithmetic is a fundamental concept in number theory that relates to the idea of divisibility and remainders. When we say that two numbers, a and b, are congruent modulo n, denoted by \( a \equiv b \mod n \), we mean that a and b leave the same remainder when divided by n. In simpler terms, n divides the difference between a and b without leaving any remainder.

Let's consider the example where we are looking for a number \( x \) that, when divided by 2, leaves a remainder of 1. This is expressed as \( x \equiv 1 \mod 2 \). Solving this means finding all numbers that are odd, since odd numbers are exactly those that leave a remainder of 1 when divided by 2. With congruence arithmetic, we can systematically solve for variables within these kinds of relationships, enabling us to work with unknowns just as we would in standard algebra, but under the additional structure imposed by the modulus.
Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is an ancient algorithm that allows for the solving of systems of linear congruences with different moduli. The theorem states that if we have several linear congruences with coprime moduli, then there is a unique solution modulo the product of these moduli.

For example, consider the congruences \( x \equiv 1 \mod 2 \) and \( x \equiv 2 \mod 3 \). Since the moduli 2 and 3 are coprime (they have no common divisor other than 1), the CRT assures us there is a unique solution for \( x \) modulo 6, which is the product of the moduli. By systematically applying the CRT to a set of linear congruences, we can determine an overall solution that satisfies all individual congruences within a system.
Modular Arithmetic
Modular arithmetic, often referred to as 'clock arithmetic', is a system of arithmetic for integers where numbers 'wrap around' after reaching a certain value, known as the modulus. It's a way of considering integers within a finite set of numbers, from 0 to one less than the modulus.

For instance, on a 12-hour clock, after 12 o'clock comes 1 o'clock again. Similarly, in modular arithmetic, operations like addition and multiplication continue to apply, but the results are taken 'modulo n', the modulus. In our textbook exercise and solution, modular arithmetic is the framework used to find a single integer that can substitute into each of the given congruences and satisfy all of them simultaneously. This method of calculation is essential in fields like cryptography, computer science, and even in solving standard mathematical problems.
System of Linear Congruences
A system of linear congruences consists of multiple congruences that we need to satisfy at the same time. Typically, each congruence will have a different modulus, and the goal is to find a number or a set of numbers that work for all congruences in the system.

To solve a system of linear congruences, we can often use a step-by-step approach. By initially solving each congruence individually and then progressively combining pairs of congruences, as demonstrated in the textbook solution, we narrow down the possibilities. The challenge lies in considering all the different remainders and moduli; however, by employing methods like the Chinese Remainder Theorem, which ties in closely with modular arithmetic and congruence arithmetic, we can systematically approach such systems and uncover solutions.

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Most popular questions from this chapter

a) Determine whether each of the following pairs of integers is congruent modulo 8 . i) 62,118 ii) \(-43,-237\) iii) \(-90,230\) b) Determine whether each of the following pairs of integers is congruent modulo \(9 .\) i) 76,243 ii) \(-137,700\) iii) \(-56,-1199\)

Prove that in any list of \(n\) consecutive integers, one of the integers is divisible by \(n\).

A band of 17 pirates captures a treasure chest full of (identical) gold coins. When the coins are divided up into equal numbers, three coins remain. One pirate accuses the distributor of miscounting and kills him in a duel. As a result, the second time the coins are distributed, in equal numbers, among the 16 surviving pirates, there are 10 coins remaining. An argument erupts and leads to gun play, resulting in the demise of another pirate. Now when the coins are divided up, in 15 equal piles, there are no remaining coins. What is the smallest number of coins that could have been in the chest?

How many units and how many (proper) zero divisors are there in (a) \(\mathbf{Z}_{17}\) ? (b) \(\mathbf{Z}_{117} ?\) (c) \(\mathbf{Z}_{1117}\) ?

a) Find all subrings of \(\mathbf{Z}_{12}, \mathbf{Z}_{18}\), and \(\mathbf{Z}_{24}\) b) Construct the Hasse diagram for each of these collections of subrings, where the partial order arises from set inclusion. Compare these diagrams with those for the set of positive divisors of \(n(n=12 ; 18 ; 24)\), where the partial order now comes from the divisibility relation. c) Find the formula for the number of subringsin \(\mathbf{Z}_{n}, n>1\).

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