Is The Sum Of Sixth Powers Divisible By 7?

Hey guys, so I was recently grinding through an entrance exam, and a pretty neat number theory question popped up. It was a true/false situation, asking whether the following sum is divisible by $7$ for all positive integers $n$:

S = 1^{6n} + 2^{6n} + 3^{6n} + inom{ ext{dots}}{} + 2025^{6n}

Now, at first glance, this might look a little intimidating with that $6n$ exponent and the large upper limit of the summation. But, as we'll dive into, understanding a bit of modular arithmetic and Fermat's Little Theorem is all you really need to crack this one. We're going to break down why this statement holds true (or false, but spoiler: it's true!). So, grab your thinking caps, and let's get into the nitty-gritty of divisibility rules and sums of powers. We'll explore how the properties of exponents and remainders can simplify seemingly complex problems into something quite elegant. Get ready to flex those number theory muscles, because this is a classic example of how powerful abstract concepts can be when applied to concrete problems. We'll be focusing on the number $7$ as our divisor and examining the behavior of powers modulo $7$. The specific upper limit, $2025$, also plays a role, and we'll uncover how it interacts with our divisor. This isn't just about solving one problem; it's about equipping you with the tools to tackle similar problems you might encounter in your own math adventures. So, let's roll up our sleeves and get started on this fascinating journey into the world of number theory, where patterns emerge and seemingly random numbers reveal their hidden structures.

Unpacking the Problem: Powers, Sums, and Divisibility by 7

Alright, let's talk about the core of the problem, guys. We're dealing with a sum of terms, where each term is a perfect sixth power raised to the power of $n$. That is, we have $k^{6n}$ for $k$ ranging from $1$ to $2025$. The big question is whether this entire sum, $S = \sum_{k=1}^{2025} k^{6n}$, is always divisible by $7$, no matter what positive integer $n$ we choose. Divisible by $7$ simply means that when you divide the sum $S$ by $7$, the remainder is $0$. In the language of modular arithmetic, we want to know if $S \equiv 0 \pmod{7}$ for all $n \geq 1$. The fact that the exponent is $6n$ is a huge clue here. Why $6$? Why not $5n$ or $7n$? This specific exponent strongly hints that we should be looking at the properties of numbers modulo $7$. Remember, when we talk about divisibility by $7$, we're essentially classifying numbers based on their remainders when divided by $7$. These remainders can only be $0, 1, 2, 3, 4, 5,$ or $6$. The structure of powers modulo $7$ is where the magic happens. The exponent $6n$ suggests we should invoke something like Fermat's Little Theorem, which is a cornerstone in elementary number theory. It deals with the remainders of powers when divided by a prime number. Since $7$ is a prime number, Fermat's Little Theorem is our best friend here. It tells us that for any integer $a$ not divisible by $7$, we have $a^{7-1} \equiv a^6 \equiv 1 \pmod{7}$. This is a powerful statement! It means that for any number $a$ (that's not a multiple of $7$), its sixth power will always leave a remainder of $1$ when divided by $7$. And what about numbers that are divisible by $7$? Well, if $a$ is a multiple of $7$, then $a \equiv 0 \pmod{7}$, and consequently, $a^{6n} \equiv 0^{6n} \equiv 0 \pmod{7}$ (as long as $6n less 0$, which it is since $n less 0$). So, we have two cases for any integer $k$: either $k$ is a multiple of $7$, or it's not. In both scenarios, $k^6 \equiv 0 \pmod{7}$ or $k^6 \equiv 1 \pmod{7}$. Now, let's look at $k^{6n}$. We can rewrite this as $(k^6)^n$. Using our findings from Fermat's Little Theorem:

• If $k$ is a multiple of $7$, then $k^6 \equiv 0 \pmod{7}$. So, $k^{6n} = (k^6)^n \equiv 0^n \equiv 0 \pmod{7}$.
• If $k$ is not a multiple of $7$, then $k^6 \equiv 1 \pmod{7}$. So, $k^{6n} = (k^6)^n \equiv 1^n \equiv 1 \pmod{7}$.

This is absolutely key! It tells us that every term in our sum $S$, when considered modulo $7$, will either be $0$ or $1$. The term $k^{6n}$ modulo $7$ depends only on whether $k$ is divisible by $7$. This significantly simplifies the problem. Instead of dealing with potentially huge numbers, we're just looking at $0$s and $1$s. The challenge now becomes figuring out how many terms are $0 \% 7$ and how many are $1 \% 7$ within the range of $1$ to $2025$. This leads us to analyze the distribution of multiples of $7$ up to $2025$.

Fermat's Little Theorem: Our Secret Weapon

Okay, let's really hammer home why Fermat's Little Theorem is so crucial here, guys. This theorem, attributed to the brilliant Pierre de Fermat, is one of the most fundamental results in number theory, especially when dealing with prime moduli. It states that if $p$ is a prime number, then for any integer $a$, the number $a^p - a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is written as:

$$a^p \equiv a \pmod{p}$$

Now, a very useful corollary of this theorem is obtained when $a$ is not divisible by $p$. In that case, we can divide both sides of the congruence $a^p \equiv a \pmod{p}$ by $a$ (since $a$ and $p$ are coprime, this division is valid in modular arithmetic). This gives us:

$$a^{p-1} \equiv 1 \pmod{p}$$

In our specific problem, the prime modulus is $p=7$. Therefore, according to Fermat's Little Theorem, for any integer $a$ that is not a multiple of $7$, we have:

$$a^{7-1} \equiv a^6 \equiv 1 \pmod{7}$$

This is precisely what we used in the previous section, but it's worth emphasizing. This result tells us that the sixth power of any number not divisible by $7$ leaves a remainder of $1$ when divided by $7$. What if $a$ is divisible by $7$? Then $a \equiv 0 \pmod{7}$. In this case, $a^6 \equiv 0^6 \equiv 0 \pmod{7}$.

So, for any integer $k$ (whether it's a multiple of $7$ or not), we know that $k^6$ will be congruent to either $0$ or $1$ modulo $7$. Specifically:

• If $k \equiv 0 \pmod{7}$ (i.e., $k$ is a multiple of $7$), then $k^6 \equiv 0 \pmod{7}$.
• If $k \not\equiv 0 \pmod{7}$ (i.e., $k$ is not a multiple of $7$), then $k^6 \equiv 1 \pmod{7}$.

Now, let's consider our term $k^{6n}$. We can rewrite this as $(k^6)^n$.

• If $k$ is a multiple of $7$, then $k^6 \equiv 0 \pmod{7}$. So, $k^{6n} = (k^6)^n \equiv 0^n \pmod{7}$. Since $n$ is a positive integer, $6n$ is a positive exponent, and $0^n = 0$. Thus, $k^{6n} \equiv 0 \pmod{7}$.
• If $k$ is not a multiple of $7$, then $k^6 \equiv 1 \pmod{7}$. So, $k^{6n} = (k^6)^n \equiv 1^n \pmod{7}$. And $1^n = 1$ for any positive integer $n$. Thus, $k^{6n} \equiv 1 \pmod{7}$.

This is the critical insight! The value of $k^{6n} \pmod{7}$ depends only on whether $k$ is a multiple of $7$. This dramatically simplifies the sum $S$. Instead of dealing with arbitrary powers, we are looking at terms that are either $0$ or $1$ modulo $7$. The entire sum's divisibility by $7$ now hinges on how many terms in the range $1$ to $2025$ are multiples of $7$, and how many are not. Fermat's Little Theorem transforms a potentially messy problem about large exponents into a counting problem based on simple remainders. It's a beautiful demonstration of how abstract mathematical principles can provide elegant solutions to concrete problems. The structure imposed by the prime modulus $7$ and the exponent $6$ (which is $p-1$) is the key that unlocks the entire problem.

Counting the Multiples of 7

So, we've established that for our sum $S = \sum_{k=1}^{2025} k^{6n}$, each term $k^{6n}$ is congruent to $0 \pmod{7}$ if $k$ is a multiple of $7$, and congruent to $1 \pmod{7}$ if $k$ is not a multiple of $7$. To find the sum $S \pmod{7}$, we need to know how many numbers between $1$ and $2025$ are multiples of $7$, and how many are not. Let $N = 2025$. We need to count the numbers $k$ in the range $1 \leq k \leq N$ such that $k \equiv 0 \pmod{7}$ and the numbers $k$ such that $k \not\equiv 0 \pmod{7}$.

To find the number of multiples of $7$ up to $2025$, we can use integer division. The number of multiples of $7$ is $\lfloor \frac{N}{7} \rfloor$. Let's calculate this:

$$\lfloor \frac{2025}{7} \rfloor$$

Performing the division: $2025 \div 7$. $2025 = 7 \times 289 + 2$. So, $\frac{2025}{7} = 289$ with a remainder of $2$. Therefore, $\lfloor \frac{2025}{7} \rfloor = 289$. This means there are exactly $289$ multiples of $7$ in the set {1, 2, lockquote , 2025}. These are the numbers 7, 14, 21, lockquote , 7 \times 289 = 2023.

For these $289$ numbers, the term $k^{6n}$ will be congruent to $0 \pmod{7}$.

Now, how many numbers are not multiples of $7$ in this range? The total number of integers from $1$ to $2025$ is $2025$. If $289$ of them are multiples of $7$, then the number of integers that are not multiples of $7$ is:

$$2025 - 289$$

Let's calculate this:

$$2025 - 289 = 1736$$

So, there are $1736$ numbers between $1$ and $2025$ that are not multiples of $7$. For each of these $1736$ numbers, the term $k^{6n}$ will be congruent to $1 \pmod{7}$.

Now we can find the sum $S$ modulo $7$. The sum $S$ is the sum of $k^{6n}$ for all $k$ from $1$ to $2025$. We can group the terms based on whether $k$ is a multiple of $7$ or not:

$$S = \sum_{k=1}^{2025} k^{6n} = \sum_{k \equiv 0 \pmod{7}, 1 \leq k \leq 2025} k^{6n} + \sum_{k \not\equiv 0 \pmod{7}, 1 \leq k \leq 2025} k^{6n}$$

Taking this equation modulo $7$:

$$S \equiv \sum_{k \equiv 0 \pmod{7}, 1 \leq k \leq 2025} (0) + \sum_{k \not\equiv 0 \pmod{7}, 1 \leq k \leq 2025} (1) \pmod{7}$$

The first sum is simply a sum of zeros, so it's $0$. The second sum consists of $1736$ terms, each equal to $1$. So, the second sum is $1736 \times 1 = 1736$.

Therefore, we have:

$$S \equiv 0 + 1736 \pmod{7}$$

$$S \equiv 1736 \pmod{7}$$

Now, we need to find the remainder of $1736$ when divided by $7$. We can do this by dividing $1736$ by $7$:

$1736 \div 7$

$1736 = 7 \times 248 + 0$

So, $1736 \equiv 0 \pmod{7}$.

This means that $S \equiv 0 \pmod{7}$. The sum $S$ is indeed divisible by $7$ for all positive integers $n$. The counting process, combined with the power of Fermat's Little Theorem, reveals the answer.

Conclusion: The Statement is True!

So, there you have it, guys! The integer 1^{6n} + 2^{6n} + 3^{6n} + inom{ ext{dots}}{} + 2025^{6n} is divisible by $7$ for all positive integers $n$. We successfully navigated the problem by leveraging Fermat's Little Theorem, which is your absolute best friend when dealing with powers and prime divisors. We saw that $k^6 \equiv 1 \pmod{7}$ for any $k$ not divisible by $7$, and $k^6 \equiv 0 \pmod{7}$ for any $k$ divisible by $7$. This crucial result simplified each term $k^{6n}$ to either $0$ or $1$ modulo $7$, depending solely on whether $k$ was a multiple of $7$.

The subsequent step was a straightforward counting exercise. We determined there are $289$ multiples of $7$ between $1$ and $2025$. For these numbers, $k^{6n} \equiv 0 \pmod{7}$. The remaining $2025 - 289 = 1736$ numbers are not multiples of $7$, and for these, $k^{6n} \equiv 1 \pmod{7}$.

Summing these up modulo $7$, we get:

$$S \equiv (289 \times 0) + (1736 \times 1) \pmod{7}$$

$$S \equiv 0 + 1736 \pmod{7}$$

$$S \equiv 1736 \pmod{7}$$

Finally, we found that $1736$ is perfectly divisible by $7$ ($1736 = 7 \times 248$). Therefore, $S \equiv 0 \pmod{7}$.

This means the original statement was TRUE. It's pretty amazing how a seemingly complex expression can be resolved with a solid understanding of basic number theory principles. The choice of the exponent $6n$ was no accident; it was a direct invitation to use Fermat's Little Theorem. Always keep an eye out for such clues in math problems! It’s a testament to the elegance and interconnectedness of mathematics that such patterns exist and can be exploited to solve problems. So, next time you see a sum involving powers and a prime modulus, remember this trick: check out Fermat's Little Theorem and see if the exponent relates to $p-1$. It might just save you a lot of headache!

Keep practicing, keep questioning, and keep enjoying the beauty of numbers, folks!