Understanding the Divisibility Problem

The question asks us to find which of the given numbers divides the expression \(121^n - 25^n + 1900^n - (-4)^n\) for any natural number \(n\).

The expression involves terms raised to the power of \(n\). We can use properties of divisibility for terms of the form \(a^n \pm b^n\). A key property states that for any natural number \(n\):

• \(a^n - b^n\) is always divisible by \((a-b)\).
• \(a^n + b^n\) is divisible by \((a+b)\) if \(n\) is odd.

We can group the terms in the given expression strategically or use modular arithmetic to determine divisibility.

Applying Modular Arithmetic for Divisibility

Let the given expression be \(E = 121^n - 25^n + 1900^n - (-4)^n\).

We will check for divisibility by the factors of the options provided, or by numbers suggested by the bases in the expression. Let's analyze the expression modulo potential divisors.

Checking Divisibility by 125

Let's look at the bases modulo 125:

• \(121 \equiv -4 \pmod{125}\)
• \(25 \equiv 25 \pmod{125}\)
• \(1900 = 15 \times 125 + 25 \equiv 25 \pmod{125}\)
• \(-4 \equiv -4 \pmod{125}\)

Now, substitute these congruences into the expression \(E\):

This shows that the expression \(121^n - 25^n + 1900^n - (-4)^n\) is divisible by 125 for all natural numbers \(n\).

Checking Divisibility by 16

Let's look at the bases modulo 16:

• \(121 = 7 \times 16 + 9 \equiv 9 \pmod{16}\)
• \(25 = 1 \times 16 + 9 \equiv 9 \pmod{16}\)
• \(1900 = 118 \times 16 + 12 \equiv 12 \pmod{16}\)
• \(-4 \equiv -4 \pmod{16}\)

Now, substitute these congruences into the expression \(E\):

Now consider the term \(12^n - (-4)^n\). This is in the form \(a^n - b^n\) where \(a=12\) and \(b=-4\). According to the divisibility property, \(a^n - b^n\) is divisible by \(a-b\).

Thus, \(12^n - (-4)^n\) is divisible by 16 for all natural numbers \(n\).

Therefore, \(E \equiv 0 \pmod{16}\). The expression \(121^n - 25^n + 1900^n - (-4)^n\) is divisible by 16 for all natural numbers \(n\).

Finding the Combined Divisor

We have shown that the expression is divisible by both 125 and 16.

To find a number that must divide the expression, we can find the Least Common Multiple (LCM) of 125 and 16. First, let's find the Greatest Common Divisor (GCD) of 125 and 16.

The prime factorization of \(125 = 5^3\).

The prime factorization of \(16 = 2^4\).

Since there are no common prime factors, the GCD(125, 16) = 1. This means 125 and 16 are coprime.

The LCM of two coprime numbers is their product.

Let's calculate the product:

So, LCM(125, 16) = 2000.

Since the expression is divisible by both 125 and 16, and they are coprime, the expression must be divisible by their LCM, which is 2000.

Conclusion and Option Check

The expression \(121^n - 25^n + 1900^n - (-4)^n\) is divisible by 2000 for all natural numbers \(n\).

Let's check the given options:

• 1904
• 2000
• 2002
• 2006

The number 2000 is one of the options.

Therefore, the expression is divisible by 2000.

Revision Table: Key Concepts

Additional Information: Generalization and Properties

The problem demonstrates a powerful technique using modular arithmetic. By examining the expression modulo different numbers, we can deduce divisibility properties.

In this case, we observed that the bases have specific relationships modulo 125 and modulo 16:

• Modulo 125: \(121 \equiv -4\), \(1900 \equiv 25\). The expression becomes \(a^n - b^n + b^n - a^n\) type structure modulo 125.
• Modulo 16: \(121 \equiv 9\), \(25 \equiv 9\), \(1900 \equiv 12\), \(-4 \equiv -4\). The expression becomes \(c^n - c^n + d^n - e^n\) type structure modulo 16, where \(d^n - e^n\) is divisible by \(d-e = 12 - (-4) = 16\).

This suggests looking for numbers that can be formed by sums or differences of the bases (or their close relatives) in different combinations.

For instance, consider the grouping \((121^n - 25^n) + (1900^n - (-4)^n)\). This is divisible by \((121-25)\) and \((1900-(-4))\), i.e., by 96 and 1904. Both 96 and 1904 are divisible by 16 (\(96=16 \times 6\), \(1904 = 16 \times 119\)), confirming divisibility by 16.

Consider the grouping \((121^n - (-4)^n) + (1900^n - 25^n)\). This is divisible by \((121-(-4))\) and \((1900-25)\), i.e., by 125 and 1875. Both 125 and 1875 are divisible by 125 (\(1875 = 125 \times 15\)), confirming divisibility by 125.

Since the expression is divisible by both 16 and 125, and GCD(16, 125)=1, it must be divisible by \(16 \times 125 = 2000\).