Question

For \(\mathrm{n} \in \mathrm{N}, 2^{y^{n}}+1\) is divisible by (1) \(3^{n+11}\) (2) \(3^{n-11}\) (3) \(3^{n+1}\) (4) \(3^{n+111}\)

Short answer

Answer: The expression \(2^{y^n} + 1\) is divisible by option (3) \(3^{n+1}\).

Step-by-step solution

Rewrite the expression

First, rewrite the expression \(2^{y^n} + 1\) using laws of exponentiation. Since we are given \(\mathrm{n} \in \mathrm{N}\), write the expression as \(2^{y^n} + 1 = (2^{y^n} + 1) \cdot 1\).

Test each option

We will now test each option and see which one satisfies the divisibility condition: (1) \(3^{n+11}\): Consider the expression \((2^{y^n} + 1) \cdot 1 \pmod{3^{n+11}}\). Unfortunately, we cannot determine the divisibility without more information about \(y\). So, we cannot conclude that \(2^{y^n} + 1\) is divisible by \(3^{n+11}\). (2) \(3^{n-11}\): Consider the expression \((2^{y^n} + 1) \cdot 1 \pmod{3^{n-11}}\). Unfortunately, we cannot determine the divisibility without more information about \(y\). So, we cannot conclude that \(2^{y^n} + 1\) is divisible by \(3^{n-11}\). (3) \(3^{n+1}\): Notice that when \(n=1\), we get the expression as \(2^{y^1} + 1\). Now, if \(y\) is odd, \(2^{y^1} \equiv -1 \pmod{9}\). In this case, \(2^{y^1} + 1 \equiv 0 \pmod{9}\), which means that \(2^{y^n} + 1\) is divisible by \(3^{n+1}\). We can also verify this for other values of \(n\). So we can conclude that \(2^{y^n} + 1\) is divisible by \(3^{n+1}\). (4) \(3^{n+111}\): Consider the expression \((2^{y^n} + 1) \cdot 1 \pmod{3^{n+111}}\). Unfortunately, we cannot determine the divisibility without more information about \(y\). So, we cannot conclude that \(2^{y^n} + 1\) is divisible by \(3^{n+111}\).

Conclusion

Based on the tests above, we can conclude that for \(\mathrm{n} \in \mathrm{N}\), the expression \(2^{y^n} + 1\) is divisible by option (3) \(3^{n+1}\).

Key concepts

Laws of Exponentiation

Exponentiation laws are rules that help simplify expressions involving powers. Understanding these laws is essential when working with equations or expressions where exponents are present.
- **Product of Powers**: When multiplying two powers with the same base, add the exponents. For example, \( a^m \times a^n = a^{m+n} \).- **Power of a Power**: To raise a power to another power, multiply the exponents. For example, \( (a^m)^n = a^{m \times n} \).- **Power of a Product**: To apply a power to a product, distribute the power to each factor in the product. For example, \( (ab)^n = a^n \times b^n \).
Understanding these rules allows for simplification, making expressions easier to manage and solve. In the exercise, knowing how to handle the exponent in \(2^{y^n}\) helps determine divisibility with various powers of 3.

Modular Arithmetic

Modular arithmetic deals with integers and a specific divisor. It is also known as 'clock arithmetic' because the results "wrap around" upon reaching a certain value—the modulus. This technique simplifies computations by making them manageable within a bounded range.
- **Modulo operation**: If you have an equation \(a \equiv b \pmod{m}\), it means the remainder of \(a\) divided by \(m\) is the same as that of \(b\).- **Useful Property**: If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), the sum \(a+c \equiv b+d \pmod{m}\), and the product \(ac \equiv bd \pmod{m}\).
In the context of the original exercise, modular arithmetic is used to check divisibility conditions, like with \(2^{y^n} + 1\) and powers of 3. It ensures calculations stay within bounds while determining divisibility without performing division directly.

Base Powers

The concept of base powers comes into play when dealing with expressions like \(2^{y^n}\) in the exercise. A base power is simply a number raised to an exponent.
- **Understanding Base and Exponent**: In an expression \(b^e\), \(b\) (the base) is multiplied by itself \(e\) times.- **Exponential Growth**: As exponents increase, the result of base powers grows exponentially—a small increase in the exponent leads to a large increase in value.- **Applications**: Base powers are often used in scientific calculations, computer science, and finance.
In the divisibility test from our exercise, the expression \(2^{y^n}\) demonstrates how changing the exponent impacts the remainder when dividing by a power of another number (like 3). Understanding the base power helps when predicting outcomes with different bases like 3, as in the expressions \(3^{n+1}\).

Natural Numbers

Natural numbers are the set of positive integers starting from 1 and moving upwards (1, 2, 3, ...). These numbers are foundational in many areas of mathematics because they represent counting numbers.
- **Inclusion**: Traditionally, natural numbers include all non-negative integers: \(\mathbb{N} = \{0, 1, 2, 3, \ldots\}\) or sometimes just positive non-zero numbers.- **Properties**: These numbers are closed under addition and multiplication, meaning the sum or product of two natural numbers is also a natural number.- **Uses**: They are used in everyday counting and ordering things.
In the provided exercise, the use of natural numbers \(n\) signifies that we are dealing with integer values starting from 1, which determines valid exponents and base power scenarios when checking divisibility across different powers of 3.