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Chapter 7: Problem 9

Use Euler's theorem to evaluate \(2^{100000}(\bmod 77)\).

Short Answer

Expert verified
2^{100000} \equiv 4 \pmod{77}.

Step by step solution

01

Understand Euler's Theorem

Euler's theorem states that if \(a\) and \(n\) are coprime integers, then \(a^{\phi(n)} \equiv 1 \pmod{n}\), where \(\phi(n)\) is Euler's totient function of \(n\). This means that to solve \(2^{100000} \mod 77\), we first need to ensure that 2 and 77 are coprime, which they are since 77 is composed of prime factors 7 and 11 and 2 is not a factor of 77.
02

Calculate Euler's Totient Function \(\phi(77)\)

77 can be factored into its prime components as \(7 \times 11\). Using the formula for Euler's totient function \(\phi(n) = n \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\)...\(n \left(1 - \frac{1}{p_k}\right)\), we get:\[\phi(77) = 77 \left(1 - \frac{1}{7}\right)\left(1 - \frac{1}{11}\right) = 77 \times \frac{6}{7} \times \frac{10}{11} = 60\].
03

Use Euler's Theorem

Since \(2^{60} \equiv 1 \pmod{77}\), we can deduce that every power of 60 will also satisfy this equivalence. Thus,\[2^{100000} = (2^{60})^{1666} \times 2^{40}\]This simplifies the problem to evaluating \(2^{40} \mod 77\).
04

Simplify \(2^{40} \mod 77\)

To calculate \(2^{40} \mod 77\), start by finding lower powers of 2 modulo 77 to help:- \(2^1 \equiv 2 \pmod{77}\)- \(2^2 \equiv 4 \pmod{77}\)- \(2^4 \equiv 16 \pmod{77}\)- \(2^8 \equiv 47 \pmod{77}\)- \(2^{16} \equiv 36 \pmod{77}\)- \(2^{32} \equiv 64 \pmod{77}\)Now multiply the appropriate powers:\[2^{40} = 2^{32} \times 2^8 \equiv 64 \times 47 \equiv 4 \pmod{77}\]
05

Conclude the Final Result

Thus, applying all the breakdown steps, the value of \(2^{100000} \mod 77\) simplifies to \(2^{40} \mod 77\). Hence, our final result is \[2^{100000} \equiv 4 \pmod{77}\]

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Key Concepts

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

Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value, known as the modulus. It is like clock arithmetic, where after 12 comes 1 again. Understanding this concept helps to solve problems involving large exponents and cumbersome numbers by simplifying calculations. Let's break it down further:
  • **Modulus:** The number at which we wrap around. For example, in modulo 12 arithmetic, 14 is equivalent to 2 because 14 wraps around by subtracting 12 twice to give 2.
  • **Congruence:** When two numbers are equivalent under a modulus, we say they are congruent. This is expressed as:\( a \equiv b \pmod{m} \) where \( m \) is the modulus.
  • **Addition, Subtraction, and Multiplication:** Basic operations work similarly to regular arithmetic, but you apply the modulus to the result. For instance, \( (13 + 8) \equiv 21 \equiv 9 \pmod{12} \).
By leveraging these principles, complex calculations, like raising numbers to large powers under a modulus, become more manageable.
Euler's Totient Function
Euler's Totient Function, denoted as \( \phi(n) \), is a crucial concept in number theory. It counts how many integers up to \( n \) are relatively prime to \( n \). This means, given \( n \), it tells us how many numbers smaller than \( n \) do not share any factors with \( n \), except for 1.
  • **Formula:** If \( n \) is a product of distinct prime numbers \( p_1, p_2, ..., p_k \), then\[ \phi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_k}\right) \]
  • **Example Calculation:** For a number like 77, which is \( 7 \times 11 \), we find \( \phi(77) = 77 \times \frac{6}{7} \times \frac{10}{11} = 60 \).
  • **Application:** This function is pivotal when using Euler鈥檚 theorem because it allows us to simplify large powers modulo \( n \).
Understanding the totient function helps us predict patterns and simplify equations in modular arithmetic.
Coprime Integers
Coprime integers, or relatively prime integers, are two numbers that have no common factors other than 1. In mathematical terms, integers \( a \) and \( b \) are coprime if their greatest common divisor (GCD) is 1.
  • **Implication:** Coprime numbers do not necessarily have to be individually prime. For example, \( 8 \) and \( 15 \) are coprime because their GCD is 1, despite neither number being prime.
  • **Significance in Euler鈥檚 Theorem:** Euler鈥檚 theorem applies to a base \( a \) and modulus \( n \) when \( a \) and \( n \) are coprime. This makes the theorem robust in applications involving different bases and moduli.
  • **Identifying Coprimes:** To check if two numbers are coprime, calculate their greatest common divisor. If it equals 1, they are coprime.
Recognizing coprime numbers is vital for applying Euler鈥檚 theorem and solving modular arithmetic problems efficiently.

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Most popular questions from this chapter

Verify the following: (a) For any positive integer \(n, \frac{1}{2} \sqrt{n} \leq \phi(n) \leq n\). [Hint: Write \(n=2^{k_{0}} p_{1}^{k_{1}} \cdots p_{r}^{k_{r}}\), so \(\phi(n)=2^{k_{0}-1} p_{1}^{k_{1}-1} \cdots p_{r}^{k_{r}-1}\left(p_{1}-1\right) \cdots\left(p_{r}-1\right) .\) Now use the inequalities \(p-1>\sqrt{p}\) and \(k-\frac{1}{2} \geq k / 2\) to obtain \(\phi(n) \geq\) \(\left.2^{k_{0}-1} p_{1}^{k_{1} / 2} \cdots p_{r}^{k_{r} / 2} .\right]\) (b) If the integer \(n>1\) has \(r\) distinct prime factors, then \(\phi(n) \geq n / 2^{r}\). (c) If \(n>1\) is a composite number, then \(\phi(n) \leq n-\sqrt{n}\). [Hint: Let \(p\) be the smallest prime divisor of \(n\), so that \(p \leq \sqrt{n}\). Then \(\phi(n) \leq n(1-1 / p) .]\)

(a) Given \(k>0\), establish that there exists a sequence of \(k\) consecutive integers \(n+1\), \(n+2, \ldots, n+k\) satisfying $$ \mu(n+1)=\mu(n+2)=\cdots=\mu(n+k)=0 $$ [Hint: Consider the following system of linear congruences, where \(p_{k}\) is the \(k\) th prime: $$ \left.x \equiv-1(\bmod 4), x \equiv-2(\bmod 9), \ldots, x \equiv-k\left(\bmod p_{k}^{2}\right) .\right] $$ (b) Find four consecutive integers for which \(\mu(n)=0\).

$$ \text { Prove that if the integer } n \text { has } r \text { distinct odd prime factors, then } 2^{r} \mid \phi(n) \text { . } $$

Show that there are infinitely many integers \(n\) for which \(\phi(n)\) is a perfect square. [Hint: Consider the integers \(n=2^{2 k+1}\) for \(k=1,2, \ldots\) ]

If \(m\) and \(n\) are relatively prime positive integers, prove that $$ m^{\phi(n)}+n^{\phi(m)} \equiv 1(\bmod m n) $$
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