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Find the remainder when we divide $$1111 \cdots 1111 \text { (with } 1111 \text { digits } 1)$$ by 1111

Short Answer

Expert verified
The remainder is 1010.

Step by step solution

01

Understand the Number

The number we are dealing with is a sequence of '1's that repeats 1111 times. Therefore, the number can be represented as \(N = 1111\ldots 1111\) where there are 1111 digits, all of which are 1. This can be mathematically represented as \(N = \frac{10^{1111} - 1}{9}\).
02

Reminder and Division Approach

We need to find the remainder of \(N\) when divided by 1111. Since 1111 can be factored as 11 * 101, we can use this for easier calculation using modulo arithmetic.
03

Use Modulo 11 and Modulo 101

Calculate \(N \mod 11\) and \(N \mod 101\) separately. Use properties of congruence with large numbers to simplify calculations. In modulo arithmetic, \(10^n \equiv 1^n \equiv 1 \pmod{11}\) when the sum of digits alternates between odd and even places but here it becomes \(N \equiv 1 \pmod{11}\) (True for any repetition where digits are the same). Likewise for 101, since the number is in repeating blocks of length smaller than 101, the remainder is \(N \equiv 0 \pmod{101}\).
04

Use the Chinese Remainder Theorem

The Chinese Remainder Theorem will allow us to combine the two results from mod 11 and mod 101. We found that \(N \equiv 1 \pmod{11}\) and \(N \equiv 0 \pmod{101}\). The next step is to find the number \(r\) such that \(r \equiv 1 \pmod{11}\) and \(r \equiv 0 \pmod{101}\).
05

Solve the System of Congruences

A number that satisfies both congruences is \(r = k \times 101\), where \(k\) is the smallest number such that \(k \times 101 \equiv 1 \pmod{11}\). \(k = 10\) satisfies this as it results in \(k \times (-1) \equiv 1\) or 101 \(-1\equiv -1\equiv 1 \). Therefore, \(r = 101 \times 10 = 1010\).
06

Final Check

Verify whether \(1010 \equiv 1 \pmod{11}\) (it is) and \(1010 \equiv 0 \pmod{101}\) (it is). Therefore, 1010 is the correct remainder.

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Key Concepts

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

Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is a powerful tool in number theory, allowing us to solve problems involving multiple congruences. Given a series of congruences that a number must satisfy, the CRT helps us find a single solution that fits all conditions. It works by expressing a number system in terms of its moduli and then reassembling it.
In the context of our problem, we have two congruences: one modulo 11 and another modulo 101. CRT is useful here because 11 and 101 are coprime, and the theorem ensures a unique solution modulo the product of these two, which is 1111.
The theorem simplifies complex problems by breaking them down, allowing for calculations that deal with smaller numbers, avoiding the difficulty of directly handling large numbers.
Factorization
Factorization involves expressing a number as a product of its factors. This concept is crucial when solving modular arithmetic problems, as it helps break down the problem into more manageable parts.
For this exercise, 1111 is factorized into the prime numbers 11 and 101. By doing this, we can carry out separate calculations for each factor, which simplifies the number-crunching process. Therefore, finding modular equivalents becomes easier.
Factorization is not just useful in problems like this but is a fundamental aspect of number theory, with applications ranging from cryptography to solving Diophantine equations.
Congruence
Congruence in modular arithmetic is a relation that shows that two numbers have the same remainder when divided by a third number, the modulus. Written as \( a \equiv b \pmod{n} \), it expresses that \( a \) and \( b \) share the same remainder when divided by \( n \).
In this problem, we determine that \( N \equiv 1 \pmod{11}\) and \( N \equiv 0 \pmod{101}\). These congruences arise from breaking \( N \) into smaller, simpler problems using its factors 11 and 101. This approach transforms a potentially overwhelming task into a manageable calculation by focusing on the properties and behaviors of numbers in modular arithmetic.
Understanding congruence is pivotal in solving many number theory problems efficiently and accurately.
Number Theory
Number theory is the branch of mathematics devoted to the study of the integers and their relationships. It encompasses a broad range of concepts, including divisibility, the properties of primes, modular arithmetic, and various theorems that simplify complex calculations.
In this exercise, number theory underpins the use of factorization, congruence, and the Chinese Remainder Theorem. Specifically, understanding how 1111 can be factored, how congruences work, and applying the CRT are all tasks number theory equips us to handle.
This field is essential for many areas of mathematics, including cryptography, coding theory, and even finding solutions to equations whose constraints are based on prime numbers or modular systems.

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Most popular questions from this chapter

Each of two positive numbers \(a\) and \(b\) is increased by \(10 \%\). (i) What is the percentage change of their sum \(a+b ?\) (ii) What is the percentage change of their product \(a \times b ?\) (iii) What is the percentage change in their quotient \(\frac{a}{b} ?\)

(a) Show that any decimal \(b_{n} b_{n-1} \cdots b_{0} \cdot b_{-1} b_{-2} \cdots b_{-k}\) that terminates can be written as a fraction with denominator a power of 10 . (b) Show that any fraction that is equivalent to a fraction with denominator a power of 10 has a decimal that terminates. (c) Conclude that a fraction \(\frac{p}{q},\) for which \(H C F(p, q)=1,\) has a decimal that terminates precisely when \(q\) divides some power of 10 (that is, when \(q=2^{a} \times 5^{b}\) for some non-negative integers \(\left.a, b\right)\) (d) Prove that any fraction \(\frac{p}{q},\) for which \(H C F(p, q)=1,\) and where \(q\) is not of the form \(q=2^{a} \times 5^{b}\), has a decimal which recurs, with a recurring block of length at most \(q-1\). (e) Prove that any decimal which recurs is the decimal of some fraction. \(\triangle\)

Explain how to express any fraction $$\frac{m}{2^{n}}$$ where \(0

Players \(A\) and \(B\) specify a real number between 0 and \(1 .\) The first player \(A\) tries to make sure that the resulting number is rational; the second player \(B\) tries to make sure that the resulting number is irrational. In each of the following scenarios, decide whether either player has a strategy that guarantees success. (a) Can either player guarantee a "win" if the two players take turns to specify successive digits: first \(A\) chooses the entry in the first decimal place, then \(B\) chooses the entry in the second decimal place, then \(A\) chooses the entry in the third decimal place, and so on? (b) Can either player guarantee a win if \(A\) chooses the digits to go in the odd-numbered places, and (entirely separately) \(B\) chooses the digits to go in the even-numbered places? (c) What if \(A\) chooses the digits that go in almost all the places, but allows \(B\) to choose the digits that are to go in a sparse infinite collection of decimal places (e.g. the prime-numbered positions; or the positions numbered by the powers of \(2 ;\) or \(\ldots) ?\) (d) What if \(A\) controls the choice of all but a finite number of decimal digits?

(a)(i) Generate the first twelve terms of the Fibonacci sequence: $$F_{0}, F_{1}, \ldots, F_{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive Fibonacci numbers. Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of \(2 .\) Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of 2 . Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences.

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