/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Compute \(2^{1000005}\) mod 55. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute \(2^{1000005}\) mod 55. Hint: This needs virtually no calculation.

Short Answer

Expert verified
\(2^{1000005} \mod 55 = 12\).

Step by step solution

01

Understanding the Problem

We need to calculate a very large exponentiation, specifically, find the remainder when \(2^{1000005}\) is divided by 55. This is the problem of finding \(2^{1000005} \mod 55\). However, we're given the hint that this requires minimal computation.
02

Use of Chinese Remainder Theorem

We can use the Chinese Remainder Theorem (CRT) to simplify our computation. Since 55 can be factored into prime factors as 5 and 11, we need to compute \(2^{1000005} \mod 5\) and \(2^{1000005} \mod 11\), and then combine these results using CRT.
03

Calculating Modulo 5

We find \(2^{1000005} \mod 5\). Observing powers of 2 modulo 5: \(2^1 \equiv 2\), \(2^2 \equiv 4\), \(2^3 \equiv 3\), \(2^4 \equiv 1\). The powers of 2 repeat every 4 terms (the order of 2 modulo 5 is 4). So, \(2^{1000005} \equiv 2^{(1000005 \mod 4)} \equiv 2^1 \equiv 2 \mod 5\).
04

Calculating Modulo 11

For \(2^{1000005} \mod 11\), the cycle length (order of 2 modulo 11) is 10 because the repeated sequence is every 10 terms: \(2^1 \equiv 2\), \(2^2 \equiv 4\), ..., \(2^{10} \equiv 1\). Hence, \(2^{1000005} \equiv 2^{(1000005 \mod 10)} \equiv 2^5 \equiv 32 \equiv 10 \mod 11\).
05

Combine Results Using CRT

Now, we have two congruences: \(2^{1000005} \equiv 2 \mod 5\) and \(2^{1000005} \equiv 10 \mod 11\). We combine them using the CRT: We need a number \(x\) such that \(x \equiv 2 \mod 5\) and \(x \equiv 10 \mod 11\). By testing, \(x = 12\) satisfies these conditions.

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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 fascinating and useful system in mathematics that revolves around the idea of "wrapping around" numbers once they reach a certain value. This value is known as the modulus. Imagine the numbers on a clock. After 12, we don't continue to 13; we wrap around and start again at 1. This is an excellent visual example of modular arithmetic in action with a modulus of 12.

With modular arithmetic, you only focus on the remainder when a number is divided by a modulus. For example, in the problem of finding \(2^{1000005} \mod 55\), we're interested in the remainder when the massive number \(2^{1000005}\) is divided by 55. This avoids the need to work with and calculate very large numbers directly.

It's essential when working with modular arithmetic to understand the concept of equivalence classes. Two numbers are equivalent under a modulus if they leave the same remainder when divided by the modulus. For instance, 8 and 3 are equivalent modulo 5 because both leave a remainder of 3 when divided by 5.
Number Theory
Number theory is one of the oldest and most intriguing branches of mathematics. It primarily deals with the properties and relationships of integers. Within number theory, there are special tools and theorems developed to handle unique number properties, like the modular arithmetic explored here.

An important concept in number theory is the prime factorization of numbers, which plays a crucial role in problems such as those solved using the Chinese Remainder Theorem (CRT). In our example, breaking down 55 into its prime factors, 5 and 11, allows us to use these simpler systems to solve complex problems.

Number theory also introduces important theorems like Fermat's Little Theorem, which tells us that if \(p\) is a prime number, then for any integer \(a\) not divisible by \(p\), \(a^{p-1} \equiv 1 \mod p\). These theorems aid in computations involving very large powers, making them manageable within modular systems.
Exponentiation in Modular Arithmetic
Exponentiation in modular arithmetic can seem daunting, especially with large numbers. However, there are elegant methods to simplify these calculations, mainly by observing patterns and using properties of cycles.

For example, when calculating \(2^{1000005} \mod 5\), instead of computing directly, we look for patterns in powers of 2 under modulo 5. This leads to the discovery of a repeating cycle, such as \(2^1 \equiv 2\), \(2^2 \equiv 4\), \(2^3 \equiv 3\), \(2^4 \equiv 1\), after which the pattern repeats. Knowing these cycles allows us to drastically simplify our computation by reducing large exponents to smaller ones, often just within the cycle.

Another helpful technique is the use of Fermat's Little Theorem, indicating that for a prime modulus \(p\), \(a^{p-1} \equiv 1 \mod p\), enabling reductions of large powers. These approaches render the calculations feasible and reliable, even with extensive numbers, as in our example.

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Most popular questions from this chapter

Let \(p \in \mathbb{N}\) be an odd prime. (i) Prove that 4 divides \(p-1\) if \(-1\) is a square modulo \(p\). Hint: Lagrange's theorem. (ii) Prove the converse of (i). Hint: Consider \(a^{(p-1) / 4}\) for a nonsquare \(a \in \mathbb{F}_{p}^{\times}\). (iii) Conclude that the Legendre symbol \(\left(\frac{-1}{p}\right)\) is 1 if and only if \(p \equiv 1 \bmod 4\).

Which of the two integers \(10^{200}+349\) and \(10^{200}+357\) is probably prime and which is certainly composite? You may use a computer algebra system to find this out, but you should not use routines like isprime or ifactor. Warning: not every exponentiation routine is suited for solving this task.

Show that for each positive integer \(n\), there exists a positive integer \(a\) such that \(a, a+1\), \(a+2, \ldots, a+n\) are all composite.

(i) Find all Carmichael numbers of the form \(3 p q\), where \(p \neq q\) are primes. (ii) Find all Carmichael numbers of the form \(5 p q\), where \(p \neq q\) are primes. (iii) Show that for any fixed prime number \(r\) there are only finitely many Carmichael numbers of the form \(r p q\), with distinct primes \(p, q\).

Let \(N \in \mathbb{N}_{\geq 3}\) be odd, \(\sigma: \mathbb{Z}_{N}^{\times} \longrightarrow \mathbb{Z}_{N}^{\times}\)the power map \(\sigma(a)=a^{(N-1) / 2}\), and \(T=\operatorname{im}(\sigma) \subseteq \mathbb{Z}_{N}^{\times}\) (i) Show that \(T=\\{1,-1\\}\) if \(N\) is prime. (ii) Prove that \(T \neq\\{1,-1\\}\) if \(N\) is not a prime power. Hint: Assume that \(-1 \in T\) and apply the Chinese Remainder Theorem. (iii) Show that \(T \neq\\{1,-1\\}\) if \(N=p^{e}\) for a prime \(p \in \mathbb{N}\) and \(e \in \mathbb{N}_{\geq 2}\). (iv) Prove that \(N\) is a Carmichael number if \(T=\\{1\\}\). (v) Consider the following algorithm.

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