# Thread: Units digit of m^5 - m is 0

1. ## Units digit of m^5 - m is 0

Dear All,

A question that I have been trying to prove ;

prove that all postive numbers (m), the units digit of m^5 - m is 0 ?

Since the integer is in base 10,
suppose m=0 then (0)^5 - 0 = 0 ( this is not a postive int)
m=1 then (1)^5 - 1 = 0
m=2 then (2)^5 - 2 = 30 ....
.......

using (10m + u)^2 is a method of proving however I have yet to prove it for this question.

regards

S.J

2. Originally Posted by srvsrv2
A question that I have been trying to prove ;
prove that all postive numbers (m), the units digit of m^5 - m is 0 ?
That is equivalent to showing $m^5-m$ is a multiple of ten.
Use induction.

3. Originally Posted by srvsrv2
Dear All,

A question that I have been trying to prove ;

prove that all postive numbers (m), the units digit of m^5 - m is 0 ?

Since the integer is in base 10,
suppose m=0 then (0)^5 - 0 = 0 ( this is not a postive int)
m=1 then (1)^5 - 1 = 0
m=2 then (2)^5 - 2 = 30 ....
.......

using (10m + u)^2 is a method of proving however I have yet to prove it for this question.

regards

S.J

$m^5-m=m(m-1)(m+1)(m^2+1)$ . It's easy now to check that this number is even and a multiple of 5 for any integer value of m.

Tonio

4. You can prove it using Fermat's Little Theorem or By mathematical induction, both works.

5. Thanks. I shall be posting my results very shortly.

6. Originally Posted by srvsrv2
Thanks. I shall be posting my results very shortly.
you can try it by recalling that $m^5$ and $m$ have the same unit digits (this is a nine-timed repetition argument)

7. Originally Posted by teachermath
you can try it by recalling that $m^5$ and $m$ have the same unit digits (this is a nine-timed repetition argument)
Assuming you mean to check each of the residue classes this is technically a "ten-timed" repetition argument.