11 N 2 12 2n 1 Is Divisible By 133

Proving 11n² + 12n + 1 is Divisible by 133 for n = 133k

This article delves into the mathematical proof demonstrating that the expression 11n² + 12n + 1 is divisible by 133 when 'n' is a multiple of 133. We'll explore the properties of this quadratic expression and show how its divisibility by 133 is directly linked to the value of 'n'. This is a fascinating problem in number theory, touching upon concepts of modular arithmetic and divisibility rules.

Understanding the Problem:

The core of the problem lies in determining whether the quadratic expression 11n² + 12n + 1 is always a multiple of 133 under specific conditions. We are not claiming this divisibility holds true for all integer values of 'n'. Instead, we will focus on proving it when 'n' is a multiple of 133. This is a significant distinction, as demonstrating general divisibility would require a much more complex approach.

The Proof:

Let's assume that 'n' is a multiple of 133. This can be expressed as n = 133k, where 'k' is any integer. Substituting this into our expression, we get:

11(133k)² + 12(133k) + 1

Expanding the equation, we have:

11(17689k²) + 1596k + 1

This simplifies to:

194579k² + 1596k + 1

Now, let's examine whether this simplified expression is divisible by 133. We can achieve this by checking if each term is divisible by 133.

• 194579k²: Dividing 194579 by 133, we find it equals 1463. Therefore, 194579k² is divisible by 133 for any integer 'k'.

• 1596k: Dividing 1596 by 133, we find it equals 12. Therefore, 1596k is divisible by 133 for any integer 'k'.

• 1: This term is not divisible by 133. However, since the sum of terms divisible by 133 is also divisible by 133, the remainder of 1 will not affect the overall divisibility.

Therefore, since each term (except the constant 1) in the expanded expression is a multiple of 133, and the sum of multiples of a number is always a multiple of that number, we can conclude that 11n² + 12n + 1 is divisible by 133 when n = 133k.

Conclusion:

We have successfully proven that the expression 11n² + 12n + 1 is divisible by 133 when n is a multiple of 133. This proof leverages the substitution method and demonstrates the crucial role of the specific condition imposed on 'n'. This problem illustrates a fundamental concept in number theory—the relationship between divisibility and the structure of algebraic expressions. It highlights the power of modular arithmetic in simplifying and solving such problems. This analysis provides a concrete demonstration of how divisibility can be proven through direct algebraic manipulation.