On the fifth power

All whole numbers can be expressed as sums of certain powers of ten; 437=4x10^2 + 3x10^1 + 7x10^0 = 400 + 30 + 7. When raised to any power all terms except the “ones” place will end with 0, since all multiples of powers of 10 end with 0.

As stated above, this works for all powers. For example, 79^2 will end with 1, since 9^2 is 81, and all other digits contribute nothing to the “ones” place. Indeed 79^2=6241.

All that’s left to show for the power of 5 is that that the pattern holds for the ten digits 0 through 9, so...

0^5=0

1^5=1

2^5=32

3^5=243

4^5=1024

5^5=3125

6^5=7776

7^5=16807

8^5=32768

9^5=59049

Fermat's Little Theorem states that, for any prime number $p$ and integer $a$ , $a^p \equiv a \pmod{p} \ \left(\Leftrightarrow p \mid a^p - a\right)$

Especially, for $p = 2$ , $a^2 \equiv a \pmod{2} \implies a^5 \equiv a \pmod{2} \implies 2 \mid a^5 - a$

and for $p = 5$ , $a^5 \equiv a \pmod{5} \implies 5 \mid a^5 - a$

Now, by the divisibility rule, for primes $p$ and $q$ , $p \mid n \wedge q \mid n \implies pq \mid n$

$2 \mid a^5 - a \wedge 5 \mid a^5 - a \implies 10 \mid a^5 - a \implies a^5 \equiv a \pmod{10}$

which is equivalent to $a$ having the same last digit as $a^5$ . $\square$

Use Fermat's theorem, $a^5$ is congruent to a(mod 5) and also 5 is odd so congruent to a (mod 10) (and not a+5 or a-5 since 10 is even and so if a is odd then the remainder it leaves on division by 10 is odd,even if a is even,so I can not add or substract 5 from a and hence is congruent to a mod 10 only. )...DONE!!!!

You can express any number as: $N = 10n+d$ ,

with $d$ a single digit number, for example: $48 = 10*4 + 8$ and $1237 = 10*123 + 7$ .

You can now compute $N^{5}$ using Pascal's Triangle coefficients for the fifth power [1, 5, 10, 10, 5, 1]:

$N^{5} = \left ( 10 n + d \right )^{5} = 10^{5}n^{5} + 5*10^{4}n^{4}d + 10*10^{3}n^{3}d^{2} + 10*10^{2}n^{2}d^{3} + 5*10nd^{4} + d^{5} =$

$10*\left ( 10^{4}n^{5} + 5*10^{3}n^{4}d + 10^{3}n^{3}d^{2} + 10^{2}n^{2}d^{3} + 5nd^{4}\right ) + d^{5}$

As you can easily see, each member of the polynomial is multiplied by 10 except for the last $d^{5}$ , and this means that the last digit of the result depends only from it. So if we compute each fifth power for every possible value of $d$ (from 0 to 9), we can verify that the last digit is the same as $d$ . And since $d^{5}$ is the only one that contributes to the last digit in the polynomial, we can confirm that the last digit of every whole number and that of its fifth power are always the same.

First, every whole number can be written as a sum of two parts, the last digit which we'll call $a$ and the rest including the tens place and up we can call $b$ . For example, 12 can be written as $2 + 10$ and 123 can be written as $3 + 120$ .

Thus we can write the fifth power any number greater than 9 as a binomal expression: $(a + b)^5$

Expanding this expression obtains: $a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$

We know that b is a multiple of 10, so all of the terms in our expansion involving $b$ will end in 0, not affecting our last digit.

The only term that does influence the term is then the only term without $b$ : $a^5$

Since $a$ is a whole number from 1 to 9, we now only need to prove for whole numbers 1 through 9!

And it's true that the last digit of whole numbers 1 through 9 is the same as the last digit of their fifth powers!

$5^5 = 3125$
$6^5 = 7776$
$7^5 = 16807$
$8^5 = 32768$
$9^5 = 59049$

Therefore, the last digit of a whole number and that of its fifth power are always the same.

Thanks for reading! This is my first attempt at this kind of thing so sorry if its messy, suggestions help!

All non-negative integers $n$ can be written as $n=10a+b$ , where $a$ and $b$ are non-negative integers, with $b \le 9$ or the last digit of $n$ . Then we have $n^5 \equiv (10a+b)^5 \equiv b^5 \text{ (mod 10)}$ . That is to say the last digit of $n^5$ is equal to the last digit of $b^5$ or the fifth power of its last digit $b$ .

Case 1 : For $b = 1,3,7,9$ being a coprime of 10, then Euler's theorem applies.

$\begin{aligned} n^5 & \equiv b^{5 \bmod \color{#3D99F6} \phi(10)} \text{ (mod 10)} & \small \color{#3D99F6} \text{where }\phi(10) \text{ is the Euler's totient function of 10.} \\ & \equiv b^{5 \bmod \color{#3D99F6} 4} \text{ (mod 10)} \\ & \equiv \boxed{b} \text{ (mod 10)} \end{aligned}$

Case 2 : For $b=5$ , then $n^5 \equiv 5^5 \equiv 5 \equiv \boxed{b} \text{ (mod 10)}$ , since all powers of 5 end with 5.

Case 3 : For $b=2,4,6,8$ multiples of 2. Then $n^5 \equiv b^5 \equiv 0 \text{ (mod 2)}$ and $n^5 \equiv b^{5 \bmod \phi(5)} \equiv b^{5 \bmod 4} \equiv b \text{ (mod 5)}$ $\implies n^5 \equiv 5m +b$ , where $m$ is an integer. Then we have

$\begin{aligned} 5m+b & \equiv 0 \text{ (mod 2)} & \small \color{#3D99F6} \text{Note that } b \text{ is a multiple of 2.} \\ m & \equiv 0 \text{ (mod 2)} \\ \implies n^5 & \equiv 5(0) + b \equiv \boxed{b} \text{ (mod 10)} \end{aligned}$

Therefore the answer is: Yes , for all non-negative $n$ , $n^5 \equiv \boxed{b} \text{ (mod 10)}$ , where $b$ is the last (unit) digit of $n$ .

If a number's last digit is $x$ it can be written in the form: $A + x$ , where $A$ is multiple of 10 (since its last digit is 0) and can be written in the form : $10 \times a$ (where a is an integer).

Then, if the last digit of $(A + x)^{5}$ is x, then the last digit of $(A+x)^{5}- x$ is 0, therefore it’s a multiple of 10.

$(A + x) ^5 - x = (10a + x)^5 - x = (10a)^5 + 5 \times (10a)^4 \times x + 10 \times (10a)^3 \times x^2 + 10 \times (10a)^{2} \times x^{3} + 5 \times (10a) \times x^{4} +x^{5} - x$

Let’s take only $x^5 - x$ :

$x^5 - x = x(x^4 - 1) = x \times (x^2 + 1) \times (x^2 - 1) = x \times (x^2 + 1) \times (x + 1) \times (x - 1)$

Now we can see that $x^5 - x$ is a multiple of 2 since it’s a multiple of adjecent numbers ( $x-1, x , x+1$ ). We can also see that $x^5 - x$ is a multiple of 5 since, no matter what $x$ $mod$ $5$ is, at least one factor of $x \times (x^2 + 1) \times (x + 1) \times (x - 1)$ is a multiple of 5.

For these reasons $x^5 - x$ is always a multiple of 10, so each term in $(10a)^5 + 5 \times (10a)^4 \times x + 10 \times (10a)^3 \times x^2 + 10 \times (10a)^{2} \times x^{3} + 5 \times (10a) \times x^{4} +x^{5} - x$ is a multiple of 10, which makes $(A+x)^5 - x$ a multiple of 10.

This means that the last digit of $(A + x)^5$ is always $x$ .

If $n \equiv n^5 \pmod{10}$ , then $n^5 - n = n(n-1)(n+1)(n^2+1) \equiv 0 \pmod{10}$ . One of $n$ and $n+1$ must be even, so the product must be divisible by $2$ . In addition, the product will be divisible by $5$ if:

$n \equiv 0 \pmod{5}$ $n-1 \equiv 0 \pmod{5} \implies n \equiv 1 \pmod{5}$ $n+1 \equiv 0 \pmod{5} \implies n \equiv 4 \pmod{5}$

And if $n$ is congruent to $2$ or $3$ modulo $5$ , then $n^2+1$ must be divisible by $5$ . Therefore the product must be divisible by $2$ and $5$ , so it is divisible by $10$ .

The expression $n^k \pmod {10}$ is periodic over $k$ for all digits $n$ . The digits $n$ have fundamental periods of 1,2, and 4, hence all have periods 4 hence $n \equiv n^5 \pmod {10}$

I found two ways to solve this problem.

n is a whole number.

Here is the first one, using this rule : if n ≡ a (mod k) then n^5 ≡ a^5 (mod k)

If n ≡ 0 (mod 10) then n^5 ≡ 0 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 1 (mod 10) then n^5 ≡ 1 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 2 (mod 10) then n^5 ≡ 32 ≡ 2 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 3 (mod 10) then n^5 ≡ 243 ≡ 3 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 4 (mod 10) then n^5 ≡ 1024 ≡ 4 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 5 (mod 10) then n^5 ≡ 3125 ≡ 5 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 6 (mod 10) then n^5 ≡ 7776 ≡ 6 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 7 (mod 10) then n^5 ≡ 16807 ≡ 7 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 8 (mod 10) then n^5 ≡ 32768 ≡ 8 (mod 10), so n^5 ≡ n (mod 10)

If n ≡ 9 (mod 10) then n^5 ≡ 59049 ≡ 9 (mod 10), so n^5 ≡ n (mod 10)

So n^5 ≡ n (mod 10), which means that n has the same last digit as n^5 (in base 10).

Here is the second one, using mathematical induction :

Basis : n=1, n^5 ≡ 1^5 ≡ 1 ≡ n (mod 10)

Induction : Assume true k^5 ≡ k (mod 10) Show true (k+1)^5 ≡ k+1 (mod 10)

(k+1)^5 = k^5 + 5k^4 +10k^3 + 10k^2 + 5k + 1 (k+1)^5 ≡ k + 1 + 5(k^4 + k) (mod 10)

We need to show that 5(k^4 + k) ≡ 0 (mod 10), which is equivalent to k^4 + k being an even number.

If k is an even number, then k = 2m. And k^4 + k = 16m^4 + 2m = 2(8m^4 + m) If k is an odd number, then k = 2m + 1. And k^4 + k = (2m + 1)^4 + 2m + 1 = 16m4 + 32m3 + 24m2 + 8m + 1 + 2m + 1 = 2(8m^4 + 16m^3 + 12m^2 + 5m + 1) So k^4 + k is an even number.

5(k^4 + k) ≡ 0 (mod 10), so (k+1)^5 ≡ k + 1 (mod 10)

Conclusion : n^5 ≡ n (mod 10), which means that n has the same last digit as n^5 (in base 10).

I am not that familiar with Fermat’s Little Theorem, but I don’t know if my solution have something to do with it. I will simply make an equation and test its validity.

$n^5=x +n$ where $x$ is an integer divisible by 5

This can also be rewritten as $n^5-n=x \rightarrow n(n^4-1)=x$

The first possibility is that $n$ is a number divisible by 5,then the equation will always be valid.

The second possibility is that $n$ is an odd number that is not 5 then the integer $(n^4)$ will always end in 1 so when when subtracted by 1 will end up divisible by 5 which makes the equation valid.

The third possibility is that $n$ is an even number then $(n^4)$ will end in 6 so when subtracted by 1 will also end up divisible by 5.

Graph x^5 on hp prime calculater, press Num, press Zoom-Integer.

$n \mod 10$ can only have 10 values $V=\bigl \{0,1,2,...9\bigr \}$ ; So $n^5 \mod 10 \equiv {v_i}^5 \mod 10$ where $v_i$ is one of the elements in $V$ . Check each of the 10 cases (or better 5 cases only since $10-r \mod 10 \equiv r \mod 10$ ) using calculator .

Brute force solution doing the multiplication of any number ending in N with that same number. Each row of the table shows the last digit of a number after squaring, cubing, and so forth.

For example, let's take a look at a number ending in 4. If we square a number ending in 4 the last digit will always be a 6 since 4 * 4 = 16 and we are only concerned with the last digit. Taking this number ending in 6 and multiplying it by our original term ending in 4 will give us a number ending in 4, since 6 * 4 = 24. Continue in this manner to populate the table.

Since the digit in the first column is always equal to the digit in the last column the statement holds.

The simplest of all proofs would be that there will be either 1, 2 or 4 possible numbers in the units place of a given number raised any number of times. Since the number in units place depends only on that of it's previous power last digit must repeat a cycle. Hence the period of it will be 1,2 or 4. 😊

By Euler $a^{\phi(n)}\equiv_n 1$ , so $a^{\phi(n)+1}\equiv_n a$ . With $n=10$ we have $\phi(10)=4$ , so $a^5\equiv_{10}a$ .

The last number of a product of two integers can be found by multiplying the last numbers. So is

$0*0=0, 0*0=0, 0*0=0, 0*0=0$

$1*1=1, 1*1=1, 1*1=1, 1*1=1$

$2*2=4, 4*2=8, 8*2=16, 6*2=12$

$3*3=9, 9*3=27, 7*3=21, 1*3=3$

$4*4=16, 6*4=24, 4*4=16, 6*4=24$

$5*5=25, 5*5=25, 5*5=25, 5*5=25$

$6*6=36, 6*6=36, 6*6=36, 6*6=36$

$7*7=49, 9*7=63, 3*7=21, 1*7= 7$

$8*8=64, 4*8=32, 2*8=16, 6*8=48$

$9*9=81, 1*9=9, 9*9=81, 1*9=9$ .

For all integers $\geqslant 10$ , the same applies.

1 1 1 1 1 - 1 2 4 8 6 2 - 2 3 9 7 1 3 - 3 4 6 4 6 4 - 4 5 5 5 5 5 - 5 6 6 6 6 6 - 6 7 9 3 1 7 - 7 8 4 2 6 8 - 8 9 1 9 1 9 - 9 0 0 0 0 0 - 0

The answer is yes if we prove that n^5-n= a (number that ends with a 0) so a number that can be written as... =2k.

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n^5-n=n((n^4)-1) n is either a 2k number (1) or a 2k+1 one(2) If (1), (n^4)-1 is a 2k+1 number If (2), (n^4)-1 is a 2k number

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But (a 2k number) * (a 2k+1 one) = a 2k number Hence proved

Can someone explain in simpler terms?