Thread: 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

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 is a multiple of ten.
Use induction.

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

. It's easy now to check that this number is even and a multiple of 5 for any integer value of m.

Tonio

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

Thanks. I shall be posting my results very shortly.

Originally Posted by srvsrv2

Thanks. I shall be posting my results very shortly.

you can try it by recalling that and have the same unit digits (this is a nine-timed repetition argument)

Originally Posted by teachermath

you can try it by recalling that and 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.