n^5 has the same last digit as n

[BMAD]

Assume that n is a natural number, prove that n and n⁵ will always have the same one's digit.   e.g. 13 and 13⁵=371,293 both end in 3.

[EventHorizon]

Proof 1: Modulus 10

Spoiler

The last digit is just the number mod 10.

(x * y) mod z = ((x mod z) * (y mod z)) mod z, so we just need to check x^5 mod 10 for each of 0-9.

x,^2,^3,^4,^5
0,0,0,0,0
1,1,1,1,1
2,4,8,6,2
3,9,7,1,3
4,6,4,6,4
5,5,5,5,5
6,6,6,6,6
7,9,3,1,7
8,4,2,6,8
9,1,9,1,9

Done.

Proof 2: Modulus 5

Spoiler

Let n be a natural number.  n can be represented as 5x + a, where x is an integer and a is one of {-2,-1,0,1,2} .

Case 1: a = 0

Any power of an odd multiple of 5 remains an odd multiple of 5, and must have a 5 in the ones digit.  
Any power of an even multiple of 5 remains an even multiple of 5, and must have a 0 in the evens digit.  
So n^5 has the same ones digit as n.

Case 2: a = 1 or -1

(5x +/- 1) * (5x +/- 1)
= 25x^2 +/- 10x + 1
= 1 mod 5
Squaring again it remains 1 mod 5.

Case 3: a = 2 or -2

(5x +/- 2) * (5x +/- 2)
= 25x^2 +/- 20x + 4
= 4 mod 5
From case 2, we see if we square something 4 mod 5 (a=-1), it becomes 1 mod 5.

Finishing cases 2 and 3:

n^4 = 1 mod 5, so it can be represented as 5y+1 for an integer y.
(5y+1)*(5x+a)
= 25xy + 5x + 5ay + a
= a mod 5.
So n^5 mod 5 = a mod 5 = n mod 5.
n^5 mod 2 = n mod 2 since any power of an even number remains even and any power of an odd number remains odd.
n^5 mod 10 = n mod 10, because 5 and 2 are relatively prime.
Same mod 10 = same last digit.

Done.

It's also interesting from this proof... n^4 ends in 1 for odd non multiples of 5, n^4 ends in 6 for even non multiples of 5.
Also, no fourth power of any integer ends in 2,3,4,7,8, or 9.

I have a third proof idea... I'll see if I can flush it out later.

Are you a math professor/researcher?  If not, where do you find / get inspiration for these?

[EventHorizon]

Proof 3: Factors of x^5-x

Spoiler

If x differs by x^5 by a multiple of 10, they must have the same one's digit.

x^5-x = x * (x^4-1)
= x * (x^2-1) * (x^2+1)
= x * (x-1) * (x+1) * (x^2+1)
Since x and x-1 are factors, x^5-x is even and we only need to show it is also a multiple of 5.

If x is a multiple of 5 or one away from a multiple of 5, x^5-x is a multiple of 5 since x, x-1, and x+1 are all factors.  The remaining possibilities to check differ from multiples of 5 by 2, so lets substitute them into the final factor.

x^2+1 = (5a +/- 2) * (5a +/- 2) + 1
= (25a^2 +/- 20a + 4) + 1
= 25a^2 +/- 20a +5
= 5 * (5a^2 +/- 4a + 1)

Since x^5-x will always be a multiple of 10, x and x^5 must have the same one's digit.

 

[BMAD]

On 8/25/2023 at 8:58 PM, EventHorizon said:

Proof 1: Modulus 10

  Reveal hidden contents

The last digit is just the number mod 10.

(x * y) mod z = ((x mod z) * (y mod z)) mod z, so we just need to check x^5 mod 10 for each of 0-9.

x,^2,^3,^4,^5
0,0,0,0,0
1,1,1,1,1
2,4,8,6,2
3,9,7,1,3
4,6,4,6,4
5,5,5,5,5
6,6,6,6,6
7,9,3,1,7
8,4,2,6,8
9,1,9,1,9

Done.

Proof 2: Modulus 5

  Reveal hidden contents

Let n be a natural number.  n can be represented as 5x + a, where x is an integer and a is one of {-2,-1,0,1,2} .

Case 1: a = 0

Any power of an odd multiple of 5 remains an odd multiple of 5, and must have a 5 in the ones digit.  
Any power of an even multiple of 5 remains an even multiple of 5, and must have a 0 in the evens digit.  
So n^5 has the same ones digit as n.

Case 2: a = 1 or -1

(5x +/- 1) * (5x +/- 1)
= 25x^2 +/- 10x + 1
= 1 mod 5
Squaring again it remains 1 mod 5.

Case 3: a = 2 or -2

(5x +/- 2) * (5x +/- 2)
= 25x^2 +/- 20x + 4
= 4 mod 5
From case 2, we see if we square something 4 mod 5 (a=-1), it becomes 1 mod 5.

Finishing cases 2 and 3:

n^4 = 1 mod 5, so it can be represented as 5y+1 for an integer y.
(5y+1)*(5x+a)
= 25xy + 5x + 5ay + a
= a mod 5.
So n^5 mod 5 = a mod 5 = n mod 5.
n^5 mod 2 = n mod 2 since any power of an even number remains even and any power of an odd number remains odd.
n^5 mod 10 = n mod 10, because 5 and 2 are relatively prime.
Same mod 10 = same last digit.

Done.

It's also interesting from this proof... n^4 ends in 1 for odd non multiples of 5, n^4 ends in 6 for even non multiples of 5.
Also, no fourth power of any integer ends in 2,3,4,7,8, or 9.

I have a third proof idea... I'll see if I can flush it out later.

Are you a math professor/researcher?  If not, where do you find / get inspiration for these?

I am a math professor.