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Chapter 4: Problem 22

Solve the system of congruence \(x \equiv 3(\bmod 6)\) and \(x \equiv 4(\bmod 7)\) using the method of back substitution.

Short Answer

Expert verified
The solution is \(x = 39\).

Step by step solution

01

- Understand the system of congruences

We need to solve the system of congruences: 1. \(x \equiv 3 \(\bmod 6\)\)2. \(x \equiv 4 \(\bmod 7\)\)
02

- Back substitution for congruence modulo 6

From the first congruence, we have: \(x = 6k + 3\) where \(k\) is an integer.
03

- Substitute into the second congruence

Substitute \(x = 6k + 3\) into the second congruence: \(6k + 3 \equiv 4 \(\bmod 7\)\). Simplify this to: \[6k + 3 \equiv 4 \(\bmod 7\) \Rightarrow 6k \equiv 1 \(\bmod 7\)\]
04

- Solve for k

We need to find an integer \(k\) such that \(6k \equiv 1 \(\bmod 7\)\). The inverse of 6 modulo 7 is 6, because \[6 \cdot 6 = 36 \equiv 1 \(\bmod 7\)\]. Therefore, \(k = 6\).
05

- Find the value of x

Substitute \(k = 6\) back into \(x = 6k + 3\):\[x = 6 \cdot 6 + 3 = 36 + 3 = 39\]
06

- Verify the solution

Verify the solution by checking both congruences:\[39 \equiv 3 \(\bmod 6\)\] and \[39 \equiv 4 \(\bmod 7\)\]. Both are true.

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Key Concepts

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

system of congruences
In mathematics, a system of congruences is a set of simultaneous congruence equations. In this exercise, we are given two equations:
  • \( x \equiv 3 (\bmod 6) \)
  • \( x \equiv 4 (\bmod 7) \)
Our goal is to find an integer solution for \( x \) that satisfies both equations at the same time. These systems are common in modular arithmetic and often arise in number theory.To tackle such a system, the Chinese Remainder Theorem (CRT) can sometimes be applied if certain conditions are met. Here, we use back substitution, another effective method.
modular arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers wrap around after reaching a certain value, called the modulus. For instance, in modulo 6 arithmetic:
  • \( x \equiv 9 (\bmod 6) \) simplifies to \( x \equiv 3 (\bmod 6) \) because 9 divided by 6 leaves a remainder of 3.
In our problem, we work with moduli 6 and 7:
  • First congruence: \( x \equiv 3 (\bmod 6) \)
  • Second congruence: \( x \equiv 4 (\bmod 7) \)
Understanding how to perform calculations under a given modulus is crucial for solving systems of congruences. It's like working with a clock where, instead of 12 hours, you wrap around after reaching 6 or 7.
back substitution
Back substitution is a method used to solve systems of congruences. It involves finding a solution for one congruence and substituting it into the other. Let's break down our example:
  • From \( x \equiv 3 (\bmod 6) \), we write \( x = 6k + 3 \) where \( k \) is an integer.
  • We substitute \( x = 6k + 3 \) into \( x \equiv 4 (\bmod 7) \) and simplify:
  • \( 6k + 3 \equiv 4 (\bmod 7) \rightarrow 6k \equiv 1 (\bmod 7) \)
  • Next, solve for \( k \). The multiplicative inverse of 6 modulo 7 is 6, because \( 6 \times 6 = 36 \equiv 1 (\bmod 7) \). Thus, \( k = 6 \)
  • Finally, substitute back: \( x = 6 \times 6 + 3 = 39 \)
By checking, we see \( 39 \equiv 3 (\bmod 6) \) and \( 39 \equiv 4 (\bmod 7) \), confirming our solution is correct.

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Most popular questions from this chapter

Find the smallest positive integer with exactly \(n\) different positive factors when \(n\) is $$ \begin{array}{lll}{\text { a) } 3 .} & {\text { b) } 4 .} & {\text { c) } 5} \\\ {\text { d) } 6 .} & {\text { e) } 10}\end{array} $$

Determine whether each of these integers is prime, verifying some of Mersenne's claims. $$\begin{array}{ll}{\text { a) } 2^{7}-1} & {\text { b) } 2^{9}-1} \\ {\text { c) } 2^{11}-1} & {\text { d) } 2^{13}-1}\end{array}$$

Express each nonnegative integer \(a\) less than 15 as a pair \((a \bmod 3, a \bmod 5)\)

What are the greatest common divisors of these pairs of integers? a) \(3^{7} \cdot 5^{3} \cdot 7^{3}, 2^{11} \cdot 3^{5} \cdot 5^{9}\) b) \(11 \cdot 13 \cdot 17,2^{9} \cdot 3^{7} \cdot 5^{5} \cdot 7^{3}\) c) \(23^{31}, 23^{17}\) d) \(41 \cdot 43 \cdot 53,41 \cdot 43 \cdot 53\) e) \(3^{13} \cdot 5^{17}, 2^{12} \cdot 7^{21}\) f) \(1111,0\)

How is the one's complement representation of the difference of two integers obtained from the one's complement representations of these integers?
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