why does last digit of the fifth power of a number is the same digit?

Answer 1

x is an integer.

If the last digit of x and x^5 are the same it follows that the last digit of (x^5 - x) will be 0. Proving that (x^5 - x) is divisible by 10 is equivalent to proving that the last digit of x is the same as the last digit of x^5.

x^5 - x
= x(x^4 - 1)
= x((x²)² -1²))
= x(x²+1)(x+1)(x-1)

If x is even, x will be divisible by 2
If x is odd, (x+1) will be divisible by 2

If the last digit of x is 0 or 5, x will be divisible by 5

If the last digit of x is 1 or 6, (x-1) will be divisible by 5
1 - 1 = 0
6 - 1 = 5

If the last digit of x is 2, 3, 7 or 8, (x² + 1) will be divisible by 5
2²+1 = 5
3²+1 = 10
7²+1 = 50
8²+1 = 65

If the last digit of x is 4 or 9, (x+1) will be divisible by 5
4+1 =5
9+1 = 10

x(x²+1)(x+1)(x-1) will always be divisible by 2 and 5
Therefore x(x²+1)(x+1)(x-1) will always be divisible by 2*5 = 10
(x^5 - x) will always be divisible by 10 - the last digit of (x^5 - x) is 0

For any given integer, x, the last digit of x^5 must be the same as the last digit of x

Answer 2

Just a coincidence.

In case you're interested, to prove this all you need is the last digit of the source number. Raise it to the fifth power and the last digit of the power number is the same as the last digit of the source number. This can be shown with a calculator

As I said, you only need the last digit of the source number. this is because all other digits are, at a minimum, multiplied by 10, which prevents them from affecting the last digit position (i.e. the 1's position)

Answer 3

Let the number be an integer of 2 digits.
Let it be a*10+b*1
When raised to power of 5 we get
100000*a^5+50000*a^4*b+10000*a^3*b^2+1000*a^2*b^3+50*a*b^4+b^5
Last digits of 100000*a^5, 50000*a^4*b, 10000*a^3*b^2, 1000*a^2*b^3 & 50*a*b^4 are zeros as they are multiplied by 100000, 50000, 10000, 1000 & 50 respectively
The last term is b^5 this decides the terminating digit
As b is a single digit b can take values from 0 to 9
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
As these all terminate with respective digits which are raised we can generalise this proof. This proof can be extended to any digited number.

Answer 4

Let's call the number x and break this up into 10 cases: x= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 mod (10). Then x^5 can be:

0^5 = 0 mod (10)
1^5 = 1 mod(10)
2^5 = 32 = 2 mod(10)
3^5 = 243 = 3 mod(10)
4^5 = 1024 = 4 mod(10)
5^5 = 5 mod(10) (Any power of 5 ends in a 5)
6^5 = (-4)^5 = -1024 = 6 mod(10)
7^5 = (-3)^5 = -243 = 7 mod(10)
8^5 = (-2)^5 = -32 = 8 mod(10)
9^5 = (-1)^5 = -1 = 9 mod(10)

Answer 5

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Answer 6

Any number x with a power n where n is any constant has to be divisible by the number x.

So for f(x) = x^3, for any number f(x) is always divisible by x.

As with x^5, the last digit of f(x) is always the same as the last digit of x because its the only way for f(x) to be divisible by x itself that satisfies the equation.....

So like for f(x) = x^2, if you divide f(x) by x, you will always get a real number. This goes for all constant....

f(x) = x^n where n is a constant.

f(x) / x = whole number

Answer 7

Very interesting problem. Never really thought about it. Let me think and then I will get back to you.