Proving Divisibility by 7 Using Modular Arithmetic

In this article, we will delve into the application of modular arithmetic and number theory to prove the divisibility of a large number by 7. Specifically, we will show that (1835^{1910} 1986^{2061}) is divisible by 7. We will use modular arithmetic to simplify the problem and apply Fermat's Little Theorem to reach our conclusion.

Introduction

Modular arithmetic is a fundamental concept in number theory, providing a powerful tool to analyze the properties of numbers in a modular context. This technique is particularly useful for proving divisibility and simplifying large exponentiations.

Basic Concepts

Before we dive into the main proof, let's review the key concepts:

Modular Arithmetic: An integer (a) is said to be congruent to (b) modulo (n) (written as (a equiv b pmod{n})) if (n) divides the difference (a - b). Fermat's Little Theorem: For a prime number (p) and an integer (a) such that (a) is not divisible by (p), (a^{p-1} equiv 1 pmod{p}).

Proof of Divisibility by 7

To prove that (1835^{1910} 1986^{2061}) is divisible by 7, we will first simplify the expressions modulo 7 and then combine the results using modular arithmetic.

Step 1: Simplify (1835 mod 7)

We start by finding the remainder when 1835 is divided by 7.

Calculate the division: (1835 div 7 approx 261) since (261 times 7 1827). Calculate the remainder: (1835 - 1827 8). Now, simplify (8 mod 7): (8 mod 7 1).

Thus, we have:

(1835 equiv 1 pmod{7})

Step 2: Simplify (1835^{1910} mod 7)

Using the result from step 1, we can now find (1835^{1910} mod 7).

(1835^{1910} equiv 1^{1910} equiv 1 pmod{7})

Step 3: Simplify (1986 mod 7)

We next find the remainder when 1986 is divided by 7.

Calculate the division: (1986 div 7 approx 283) since (283 times 7 1981). Calculate the remainder: (1986 - 1981 5).

Thus, we have:

(1986 equiv 5 pmod{7})

Step 4: Simplify (1986^{2061} mod 7)

Using Fermat's Little Theorem, we know that for a prime (p) and an integer (a) not divisible by (p), (a^{p-1} equiv 1 pmod{p}). Here, (p 7) and (a 5), so:

(5^6 equiv 1 pmod{7})

To find (1986^{2061} mod 7), we need to simplify the exponent 2061 modulo 6.

Calculate the division: (2061 div 6 approx 343) since (343 times 6 2058). Calculate the remainder: (2061 - 2058 3). Thus, (2061 equiv 3 pmod{6}).

Using this result, we have:

(1986^{2061} equiv 5^3 pmod{7})

Now, compute (5^3 mod 7):

(5^3 125)

(125 div 7 approx 17) since (17 times 7 119)

(125 - 119 6)

Thus, we have:

(125 equiv 6 pmod{7}) and hence ()1986^{2061} equiv 6 pmod{7})

Combining the Results

Now, we combine the results from the previous steps:

(1835^{1910} 1986^{2061} equiv 1 6 pmod{7})

(1 6 equiv 7 equiv 0 pmod{7})

Therefore, (1835^{1910} 1986^{2061}) is divisible by 7.

Conclusion

The expression (1835^{1910} 1986^{2061}) is indeed divisible by 7. The final result is:

(boxed{0})

This demonstrates the power of modular arithmetic and Fermat's Little Theorem in simplifying and solving complex divisibility problems.