Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Quick question

What is the last digit of

?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quick question

Does it help if I say that x ≡ x^5 (mod 10)?

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Why did the vector cross the road?
It wanted to be normal.

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Quick question

1+1+5+1+1=9

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You can shear a sheep many times but skin him only once.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Quick question

If x ≡ x^5 (mod 10), the value of x^n (mod 10) cycles between 4 values as n increases.

2004 is divisible by 4.

1^2004 ≡ 1^4 (mod 10) ≡ 1 (mod 10)
3^2004 ≡ 3^4 (mod 10) ≡ 1 (mod 10)
5^2004 ≡ 5^4 (mod 10) ≡ 5 (mod 10)
7^2004 ≡ 7^4 (mod 10) ≡ 1 (mod 10)
9^2004 ≡ 9^4 (mod 10) ≡ 1 (mod 10)

I can see why we use mod 10, and 1+1+5+1+1=9, but where did x ≡ x^5 (mod 10) come from in the first place? I can see it works, but why, and what made you think of it?

Thanks.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quick question

I'm not sure why it works, other than 'coincidence'.
I just thought of it because it was something I happened to know anyway.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Quick question

Euler’s theorem states that for any integers a and n which are coprime,

where the Euler function

is the number of postive integers less than n and coprime with n. We have

because 1, 3, 7, 9 are the positive integers less than and coprime with 10. Hence

where

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Quick question

Ahh I see, thanks Jane. When I was investigating a^n (mod 10) it was cycling through 1, 3, 7 and 9. I can see why now.