Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(n(>1)\) be a positive integer. Then largest integer \(m\) such that \(\left(n^{m}+1\right)\) divides \(1+n+n^{2}+\ldots+n^{255}\) is (A) 128 (B) 63 (C) 64 (D) 32

Step-by-Step Solution

Answer

The largest integer \(m\) is 128.

Step 1: Simplify the Expression

The expression given is the sum of a geometric series: \(1 + n + n^2 + \dots + n^{255}\). The formula for the sum of a geometric series is \(\frac{n^{k+1} - 1}{n-1}\), where \(k\) is the largest power, in this case, 255. Therefore, the sum is \(\frac{n^{256} - 1}{n-1}\).

Step 2: Identify Divisors

We need \(n^m + 1\) to divide \(\frac{n^{256} - 1}{n-1}\). For \(n^m + 1\) to divide \(n^{256} - 1\), it must divide \(n^{256}\) and also the difference to make up the formulation into \(n^m + 1\).

Step 3: Use Cyclotomic Polynomial Property

The expression \(n^m + 1\) divides \(n^{2m} - 1\). Therefore, we need to find the largest \(m\) such that \(m\) is a divisor of 256 and \(2m\) is the smallest multiple of the highest power that covers for \(n^{2m} - 1\) covering the total cycle of \(256\).

Step 4: Determine Suitable Value of \(m\)

Since \(n^m + 1\) should divide \(n^{256} - 1\), we try to find the largest power \(m\) dividing 256 within constraints. Since \(256 = 2^8\), possible divisors of 256 are 128, 64, 32, etc. Check for \(n^{128} + 1\), which will divide \(n^{256} - 1\), fulfilling the necessary condition.

Step 5: Verify with Divisibility

Let's take \(m = 128\). Check divisibility: \(n^{256}\) and \(n^{256} - 1\) together permit \(n^{128} + 1\) to divide properly since this conforms with \(n^{256} - 1 = (n^{128} - 1)(n^{128} + 1)\). Hence, \(n^{128} + 1\) divides, making \(m = 128\) the largest value.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The general expression for a geometric series is:
\(a + ar + ar^2 + \ldots + ar^n\).
Here, \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms minus one.
- **Sum of the Series:** The sum \(S\) of this series can be calculated using the formula:
\[ S = \frac{a(r^{n+1} - 1)}{r - 1} \] This formula helps find the total of all terms in the series quickly without adding each term separately.
- **Example:** In our given problem, \(a = 1\), \(r = n\), and \(n = 255\), thus the sum is \(\frac{n^{256} - 1}{n - 1}\). Understanding geometric series is crucial for recognizing patterns in mathematics, including finance for calculating interest, and in physics for wave patterns.

Cyclotomic Polynomial

Cyclotomic polynomials are a special class of polynomials that are closely related to complex numbers and roots of unity. They play an essential role in number theory and algebra. The \(n\)-th cyclotomic polynomial, denoted as \(\Phi_n(x)\), is defined as the product of \[ \Phi_n(x) = \prod_{1 \leq k \leq n \atop \gcd(k, n) = 1} \left(x - e^{2\pi i k/n} \right) \] where \(e^{2\pi i k/n}\) are the primitive \(n\)-th roots of unity. - **Properties:** Cyclotomic polynomials have integer coefficients and are irreducible over the rational numbers. They're used for simplifying expressions that repeat cyclically, such as powers and cycles in polynomial equations.
- **Application in Exercise:** In our context, dividing polynomials like \(n^m + 1\) relates to cyclotomic polynomials because they help understand which powers of \(n\) contribute to the divisibility criteria in problems and how factors are constructed based on gradient change.
These polynomials help simplify the understanding and solving of division problems within polynomial algebra.

Divisibility in Polynomials

Understanding divisibility in polynomials is pivotal for solving equations like division problems where one polynomial is a factor of another. This involves ensuring that one polynomial can "exactly divide" another without leaving any remainder.
- **Basic Principle:** If a polynomial \(f(x)\) divides another polynomial \(g(x)\), then there exists a polynomial \(h(x)\) such that \(g(x) = f(x)\cdot h(x)\).
- **Example:** In the given exercise, the goal was to find the largest \(m\) such that \(n^m + 1\) divides \(n^{256} - 1\), effectively making it equivalent to finding a quotient \(h(x)\).
To ensure divisibility, polynomials often align with cyclotomic polynomials, which encapsulate cycles and periodicity in polynomial order. Recognizing the correct factorization helps in simplifying complex polynomial expressions into manageable factors, thereby solving divisibility problems efficiently.