# [ASK]Show that the sum of the fifth powers of these numbers is divisible by 5

• MHB
• Monoxdifly
In summary, the sum of ten integers being 0 is just a distraction, and the key is to use Fermat's Little Theorem to show that the sum of the fifth powers of these numbers is divisible by 5. This is because for any integer, its fifth power is congruent to itself modulo 5. Therefore, the sum of the fifth powers will be congruent to the sum of the original numbers, which is 0, and thus divisible by 5.
Monoxdifly
MHB
The sum of ten integers is 0. Show that the sum of the fifth powers of these numbers is divisible by 5.

For this one I don't know what I have to do at all other than brute-forcing which may even be impossible.

Monoxdifly said:
The sum of ten integers is 0. Show that the sum of the fifth powers of these numbers is divisible by 5.

For this one I don't know what I have to do at all other than brute-forcing which may even be impossible.

Hi Monoxdifly,

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.

castor28 said:
Hi Monoxdifly,

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.

So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?

Monoxdifly said:
So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?
Hi Monoxdifly,

The distraction is the fact that there are ten integers. The fact that the sum is $0$ is essential.

More precisely, if the integers are $a, b, c, \ldots$, Fermat's theorem tells us that $a^5\equiv a, b^5\equiv b\ldots\pmod5$.

We have therefore $a^5 + b^5 + \cdots\equiv a + b + \cdots\pmod 5$. As we are told that $a + b +\cdots=0\equiv0\pmod5$, we conclude that $a^5 + b^5 + \cdots\equiv0\pmod5$, which means that the sum is divisible by $5$.

I think I kinda get a grasp here. Thanks for your help.

## 1. How do you show that the sum of the fifth powers of numbers is divisible by 5?

To show that the sum of the fifth powers of numbers is divisible by 5, we can use mathematical induction. We start by showing that the statement is true for the base case, which is when n = 1. Then, we assume that the statement is true for some arbitrary value of n, and use this assumption to prove that it is also true for n+1. This process can be repeated infinitely, proving that the statement is true for all values of n.

## 2. What is the formula for calculating the sum of the fifth powers of numbers?

The formula for calculating the sum of the fifth powers of numbers is n(n+1)(2n+1)(3n^2+3n-1)/30, where n is the number of terms in the series. This formula can be derived using the sum of consecutive cubes formula and the binomial theorem.

## 3. Can the sum of the fifth powers of numbers be divisible by 5 for any value of n?

Yes, the sum of the fifth powers of numbers can be divisible by 5 for any value of n. This is because the formula for calculating the sum of the fifth powers of numbers always results in a multiple of 5, regardless of the value of n.

## 4. How can you prove that the sum of the fifth powers of numbers is divisible by 5 using mathematical induction?

To prove that the sum of the fifth powers of numbers is divisible by 5 using mathematical induction, we need to show that the statement is true for the base case (n = 1) and then use the assumption that it is true for some arbitrary value of n to prove that it is also true for n+1. By repeating this process infinitely, we can prove that the statement is true for all values of n.

## 5. Are there any other methods for proving that the sum of the fifth powers of numbers is divisible by 5?

Yes, there are other methods for proving that the sum of the fifth powers of numbers is divisible by 5. For example, we can use modular arithmetic to show that the sum of the fifth powers of numbers is always congruent to 0 (mod 5). We can also use the properties of divisibility to show that the sum of the fifth powers of numbers is always a multiple of 5. However, mathematical induction is the most commonly used method for proving this statement.

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