Exploring Fermats Little Theorem and its Application in Finding Remainders

Introduction to Fermat's Little Theorem
Before delving into the main content, it is crucial to understand the fundamental theorem we will be utilizing - Fermat's Little Theorem. This theorem is a cornerstone in number theory and is used extensively in solving problems related to remainders and modular arithmetic. It states that if (p) is a prime number and (a) is an integer not divisible by (p), then:

[a^{p-1} equiv 1 text{ (mod } ptext{)}]

This theorem simplifies the calculation of large exponentiations modulo a prime number. Let's see how to use this theorem to find the remainder of (67^{99}) divided by 7.

Finding the Remainder of (67^{99}) Divided by 7

We begin the process by reducing the base 67 modulo 7. To do so, we perform the division:

[67 div 7 9 text{ with a remainder of } 4]

Hence,

[67 equiv 4 text{ (mod } 7text{)}]

Our goal now is to find (67^{99} mod 7). By the reduction performed, we need to calculate (4^{99} mod 7).

Using Fermat's Little Theorem

Since 7 is a prime number and 4 is not divisible by 7, we apply Fermat's Little Theorem:

[4^{6} equiv 1 text{ (mod } 7text{)}]

Next, we need to reduce the exponent 99 modulo 6 (since 6 7-1):

[99 div 6 16 text{ with a remainder of } 3]

Hence,

[99 equiv 3 text{ (mod } 6text{)}]

So we have:

[4^{99} equiv 4^{3} text{ (mod } 7text{)}]

Calculating the Result

Now, we calculate (4^{3}):

[4^{3} 64]

Next, we find 64 mod 7:

[64 div 7 9 text{ with a remainder of } 1]

Hence,

[64 equiv 1 text{ (mod } 7text{)}]

Therefore, we can conclude that:

[67^{99} equiv 4^{99} equiv 4^{3} equiv 64 equiv 1 mod 7]

The remainder of (67^{99}) when divided by 7 is 1.

Alternative Methods and Insights

Several alternative methods can be used to verify our result or attempt similar problems.

Method 1: Modular Arithmetic Reduction

We can also express 67 4 text{ (mod } 7text{)}end{p>

since 67 equiv 4 text{ (mod } 7text{)}end{p>

and by Fermat's Little Theorem, we know:

[4^{6} equiv 1 text{ (mod } 7text{)}]

Then,

[4^{99} 4^{3 cdot 33} equiv (4^{6})^{16} cdot 4^{3} equiv 1^{16} cdot 4^{3} equiv 4^{3} text{ (mod } 7text{)}]

Calculating 4^{3} end{p>

[4^{3} 64]

and

[64 mod 7 1]

So,

[67^{99} equiv 1 text{ (mod } 7text{)}]

Method 2: Using a More Direct Approach

Another way to look at this problem is to recognize some patterns in powers of 4 modulo 7:

[67 equiv 4 text{ (mod } 7text{)}]

Thus,

[67^{99} equiv 4^{99} text{ (mod } 7text{)}]

From the theorem, we know:

[4^2 equiv 2 text{ (mod } 7text{)}]

and consequently,

[4^3 equiv 4^2 cdot 4 equiv 2 cdot 4 equiv 8 equiv 1 mod 7]

Therefore,

[4^{99} equiv 1 text{ (mod } 7text{)}]

So, the remainder of 67^{99} when divided by 7 is 1.

Conclusion

Through the application of Fermat's Little Theorem, we were able to efficiently determine the remainder of 67^{99} when divided by 7. This problem showcases the power of modular arithmetic and the utility of Fermat's Little Theorem in simplifying complex calculations.