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(a) Prove that if \(d \mid n\), then \(2^{d}-1 \mid 2^{n}-1\). [Hint: Use the identity $$ \left.x^{k}-1=(x-1)\left(x^{t-1}+x^{t-2}+\cdots+x+1\right) .\right] $$ (b) Verify that \(2^{35}-1\) is divisible by 31 and 127 .

Short Answer

Expert verified
(a) Proven using identity for overall divisibility. (b) Verified that 31 and 127 divide \(2^{35}-1\).

Step by step solution

01

Understand the Relation

Given that \(d \mid n\), it means \(n = kd\) for some integer \(k\). Thus, we need to show that \(2^d - 1 \mid 2^{kd} - 1\).
02

Use the Hint

Consider the identity: \[x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + \cdots + x + 1).\] Let \(x = 2^d\), then the expression becomes \((2^d)^k - 1 = (2^d - 1)((2^d)^{k-1} + (2^d)^{k-2} + \cdots + 2^d + 1)\). This shows \(2^d - 1\) is a factor of \((2^d)^k - 1 = 2^{kd} - 1\).
03

Conclusion of Part (a)

Since \(2^d - 1\) divides \(2^{kd} - 1\), we've proved that if \(d \mid n\), then \(2^d - 1 \mid 2^n - 1\).
04

Calculate \(2^{35} - 1\)

Calculate \(2^{35} - 1\) to verify its divisibility. Calculate \(2^5 = 32\) and then subsequently calculate powers of 2 up to 35.
05

Check Divisibility by 31

Using Fermat's Little Theorem, which states \(a^{p-1} \equiv 1 \pmod{p}\) for a prime \(p\), check \(2^{30} \equiv 1 \pmod{31}\). Thus \(2^{35} \equiv 2^5 \equiv 32 \equiv 1 \pmod{31}\), confirming \(2^{35} - 1\) is divisible by 31.
06

Check Divisibility by 127

Similarly, since 127 is prime, \(2^{126} \equiv 1 \pmod{127}\). Hence \(2^{35} \equiv 1 \pmod{127}\) must also hold, implying \(2^{35} - 1\) is divisible by 127.
07

Verify the Conclusion

Both 31 and 127 divide \(2^{35} - 1\), which confirms the divisibility as required in part (b).

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Key Concepts

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

Fermat's Little Theorem
Fermat's Little Theorem is a fundamental principle in number theory that gives us a way to simplify complex calculations involving modular arithmetic. It states that if \( p \) is a prime number and \( a \) is an integer such that \( a ot\equiv 0 \pmod{p} \), then: \[ a^{p-1} \equiv 1 \pmod{p} \] This theorem is particularly useful when checking the divisibility of power expressions by a prime number. In our original exercise, we used Fermat's Little Theorem to verify the divisibility of the expression \( 2^{35} - 1 \) by the primes 31 and 127.
  • Firstly, for 31: Since \( 31 \) is prime, applying Fermat's theorem gives us \( 2^{30} \equiv 1 \pmod{31} \). Therefore, \( 2^{35} = 2^{30} \times 2^5 \equiv 1 \times 32 \equiv 1 \pmod{31} \), confirming \( 2^{35} - 1 \) is divisible by 31.

  • Secondly, for 127: Similarly, 127 being prime ensures \( 2^{126} \equiv 1 \pmod{127} \), which means \( 2^{35} \equiv 1 \pmod{127} \), thus showing \( 2^{35} - 1 \) is divisible by 127 as well.
Fermat's Little Theorem thus provides a shortcut to confirm these divisibility cases, making it an invaluable tool in number theory.
Factors of Integers
An integer is divisible by another if there is no remainder upon division. This concept plays a crucial role in understanding factorization and divisibility in mathematics. The divisibility of power expressions, such as \( 2^n - 1 \), can be investigated using identities and divisibility rules. In our exercise, we started with an identity to show that if \( d \mid n \), then \( 2^{d}-1 \mid 2^{n}-1 \). We utilized the hint provided, which is based on factoring a difference of powers:
  • The identity given was \( x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + \cdots + x + 1) \).

  • We substitute \( x = 2^d \) and \( k \) as \( n/d \), to express \((2^d)^k - 1\).

  • It reveals \( (2^d)^k - 1 = (2^d - 1)Q \), where \( Q \) is a polynomial in \( 2^d \), implying that \( 2^d - 1 \) is a factor of \( 2^{n} - 1 \).
This method allows us to see that specific structures in numbers and expressions lead to predictable patterns in divisibility.
Power of Two Expressions
Power of two expressions, like \( 2^n \), exhibit special properties that make them intriguing in mathematics. Understanding these properties requires examining how expressions grow and interact with basic arithmetic.
  • Exponential Growth: Powers of two increase exponentially, meaning each subsequent power is double the previous one. This rapid growth can lead to challenging calculations without formulas or shortcuts.

  • Divisibility Patterns: As highlighted in the original exercise, the expressions \( 2^n - 1 \) can be factored and analyzed for divisibility by specific integers. Techniques involving Fermat's theorem and polynomial identities simplify these tasks.
The original exercise showcases these properties through the divisibility of \( 2^{35} - 1 \) by 31 and 127 using these patterns. Decomposing powers of two expressions into manageable parts often involves understanding the underlying growth and using mathematical tools to predict behavior in divisibility. By analyzing patterns and using divisibility shortcuts, such tasks become manageable and intuitive.

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Most popular questions from this chapter

Use the Euclidean Algorithm to obtain integers \(x\) and \(y\) satisfying the following: (a) \(\operatorname{gcd}(56,72)=56 x+72 y\). (b) \(\operatorname{gcd}(24,138)=24 x+138 y\) (c) \(\operatorname{gcd}(119,272)=119 x+272 y\) (d) \(\operatorname{gcd}(1769,2378)=1769 x+2378 y\)

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(a) A man has \(\$ 4.55\) in change composed entirely of dimes and quarters. What are the maximum and minimum number of coins that he can have? Is it possible for the number of dimes to equal the number of quarters? (b) The neighborhood theater charges \(\$ 1.80\) for adult admissions and \(\$ .75\) for children. On a particular evening the total receipts were \(\$ 90\). Assuming that more adults than children were present, how many people attended? (c) A certain number of sixes and nines is added to give a sum of 126 ; if the number of sixes and nines is interchanged, the new sum is \(114 .\) How many of each were there originally?

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