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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

The Vigen猫re cipher is a block cipher, with a key that is a string of letters with numerical equivalents \(k_{1} k_{2} \ldots k_{m},\) where \(k_{i} \in \mathbf{Z}_{26}\) for \(i=1,2, \ldots, m .\) Suppose that the numerical equivalents of the letters of a plaintext block are \(p_{1} p_{2} \ldots p_{m} .\) The corresponding numerical ciphertext block is \(\left(p_{1}+k_{1}\right)\) mod 26 \(\left(p_{2}+k_{2}\right) \bmod 26 \ldots\left(p_{m}+k_{m}\right)\) mod \(26 .\) Finally, we translate back to letters. For example, suppose that the key string is RED, with numerical equivalents \(1743 .\) Then, the plaintext ORANGE, with numerical equivalents \(141700130604,\) is encrypted by first splitting it into two blocks 141700 and 13 \(0604 .\) Then, in each block we shift the first letter by 17 , the second by \(4,\) and the third by \(3 .\) We obtain 52103 and 0410 \(07 .\) The cipherext is FVDEKH. Use the Vigen猫re cipher with key BLUE to encrypt the message SNOWFALL.

Convert the decimal expansion of each of these integers to a binary expansion. \(\begin{array}{llll}{\text { a) } 321} & {\text { b) } 1023} & {\text { c) } 100632}\end{array}\)

Determine whether the integers in each of these sets are pairwise relatively prime. $$\begin{array}{ll}{\text { a) } 21,34,55} & {\text { b) } 14,17,85} \\\ {\text { c) } 25,41,49,64} & {\text { d) } 17,18,19,23}\end{array}$$

Give a procedure for converting from the hexadecimal expansion of an integer to its octal expansion using binary notation as an intermediate step.

Suppose that \((n, e)\) is an RSA encryption key, with \(n=p q,\) where \(p\) and \(q\) are large primes and \(\operatorname{gcd}(e,(p-1)(q-1))=1 .\) Furthermore, suppose that \(d\) is an inverse of \(e\) modulo \((p-1)(q-1) .\) Suppose that \(C \equiv M^{e}(\bmod p q) .\) In the text we showed that RSA decryption, that is, the congruence \(C^{d} \equiv M(\bmod p q)\) holds when gcd \((M, p q)=1 .\) Show that this decryption congruence also holds when \(\operatorname{gcd}(M, p q)>1 .[\text { Hint: Use con- }\) gruences modulo \(p\) and modulo \(q\) and apply the Chinese remainder theorem. \(]\)
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