Chapter 16: Problem 56
For each \(n \in N, 5^{3 n}-1\) is divisible by (1) 115 (2) 124 (3) 5 (4) 6
Short Answer
Expert verified
a) 115
b) 124
c) 248
d) 372
Answer: b) 124
Step by step solution
01
Check divisibility with 115
Recall that a number \(a\) is said to be divisible by a number \(b\) iff \(a \equiv 0\pmod{b}\). The expression we have is \(5^{3n} - 1\). We'll check its divisibility with 115 by testing \(5^{3n} \equiv 1\pmod{115}\). Substitute \(n=1\):
$$5^3 \equiv 125 \equiv 10\pmod{115}$$
It's not divisible by 115.
02
Check divisibility with 124
Test \(5^{3n} \equiv 1\pmod{124}\). Substitute \(n=1\):
$$5^3 \equiv 125 \equiv 1\pmod{124}$$
Observe that for all \(n\), we have \(5^{3n} \equiv (5^3)^n \equiv 1^n \equiv 1\pmod{124}\). Thus, \(5^{3n}-1\) is divisible by 124 for all \(n\). We can eliminate options (3) and (4).
03
Conclusion
\(5^{3n} - 1\) is divisible by 124 for all natural numbers \(n\). Thus, the correct answer is (2) 124.
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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 after reaching a certain value, which is called the modulus. This can be thought of like the hours on a clock, where after 12, the number wraps back to 1. In the context of our problem, we are checking if certain expressions are equivalent to zero modulo a number, meaning they are divisible by that number.
When we say that a number \( a \) is divisible by another number \( b \), we are essentially saying that \( a \equiv 0 \pmod{b} \). This is useful to determine divisibility because it helps simplify calculations by focusing on the remainder when divided by the modulus. For example, in our exercise, the problem is simplified using the modulo operation to check if \( 5^{3n} - 1 \) is divisible by 124, 115, 5, or 6.
When we say that a number \( a \) is divisible by another number \( b \), we are essentially saying that \( a \equiv 0 \pmod{b} \). This is useful to determine divisibility because it helps simplify calculations by focusing on the remainder when divided by the modulus. For example, in our exercise, the problem is simplified using the modulo operation to check if \( 5^{3n} - 1 \) is divisible by 124, 115, 5, or 6.
- If \( 5^{3n} - 1 \equiv 0 \pmod{124} \), it means \( 5^{3n} \equiv 1 \pmod{124} \).
- If \( 125 \equiv 1 \pmod{124} \), then this property holds for any power \( n \), making the original expression divisible by 124 for all natural numbers \( n \).
Number Theory
Number theory is a branch of pure mathematics devoted to the study of the integers and integer-valued functions. Number theory is known for its particularly wide array of concepts and problems, many of which relate to divisibility.
One key concept in number theory used in our problem is how powers of numbers behave under a modulus. In our exercise, we examine how \( 5^{3n} \) behaves modulo different integers to determine divisibility. This involves understanding congruences, which is a core concept in number theory.
One key concept in number theory used in our problem is how powers of numbers behave under a modulus. In our exercise, we examine how \( 5^{3n} \) behaves modulo different integers to determine divisibility. This involves understanding congruences, which is a core concept in number theory.
- A congruence relation states that two numbers leave the same remainder when divided by a modulus.
- Here, \( 5^3 \equiv 1 \pmod{124} \) allows us to extend this relationship to any exponent of this power, showing that \( 5^{3n} \equiv 1 \pmod{124} \) for any natural \( n \).
Exponentiation
Exponentiation refers to the process of raising a number to the power of another. It is a fundamental operation in mathematics, especially critical when it comes to modular arithmetic. In our problem, we dealt with expressions like \( 5^{3n} \) and their behavior under various moduli.
Exponentiation in the context of modular arithmetic requires us to calculate powers while considering the modulus involved. This can drastically simplify the process, especially with large numbers, as seen in this exercise.
Exponentiation in the context of modular arithmetic requires us to calculate powers while considering the modulus involved. This can drastically simplify the process, especially with large numbers, as seen in this exercise.
- For instance, instead of calculating \( 5^{3n} \) directly, we calculate \( 5^3 \equiv 1 \pmod{124} \), simplifying our original statement for all \( n \).
- This not only makes computations feasible but also helps uncover patterns, such as the repeated equivalence modulo a number, which is a form of simplification brought by exponentiation within modular arithmetic.