Math Is Fun Forum

macy

Guest

Algebra headache

can any expert here tell me how to work out the last digit of the result when six multiply with eight to the power 2009?

Thanks

juriguen

Member

Registered: 2009-07-05

Posts: 59

Re: Algebra headache

Hi!

First, lets see after how many powers does the last digit repeat:

OK, so after every 4 cycles, starting at some number N, eight to the power of N + 4k repeats its last digit.

For instance, 8^5 = 8^{1 + 4} has the same last digit as 8^1, both equal to 8.

So, for 2009

Hope it helps!
Jose

PS: there are other numbers which seem more strange, since they have 0 remainder when dividing by 4, but then N is also 4.

For example:

And, indeed, 8^20 = 1152921504606846976

Last edited by juriguen (2009-08-12 20:50:25)

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“Make everything as simple as possible, but not simpler.” -- Albert Einstein

Riad Zaidan

Guest

Re: Algebra headache

Hi macy
We can use the language of congruences as follows:
8^2=64=4(mod 10) multiply both sides withn 8^2 we get
8^4=64*4(mod 10)=256(mod 10 ) = 6(mod 10) by squaring both sides we get
8^8=36(mod 10)= 6(mod 10) so in general we have
8^(4k)=6(mod 10) where  k is a positive integer so
8^(4k+1)= 8*6(mod 10)=48(mod 10)= 8 (mod 10 )
Now for 2009 we have 2009=4*502+1 therefore
8^2009= 8^(4*502+1)= 8 (mod 10 ) multiplying both sides by 6 we get
6*(8^2009)= 6*8 (mod 10 )=48 (mod 10 ) =8 (mod 10 )
therefore the last digit is the remainder when dividing by 10 and so the remainder is 8.
Best Regards
Riad Zaidan

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra headache

Hi Macy;

Lets forget about the 6 for a moment.

Now do the inner parentheses.

8^8 ends in a 6. So  8^8 * 8^8 * 8^8 * 8^8 *8^8 * 8^8 also ends in a 6 . Now times by the last 8 =48  so 8^49 ends in 8.

Now do the outer power:

We already know that any succession of 8^8 * 8^8 * 8^8... ends in 6. The last 8 * 6 =48 which ends in 8.

So

Which is the math way of saying that 8^2009 ends in 8.

Now don't forget the 6 which is just 6*8 =48 the last digit is again an 8.

Last edited by bobbym (2009-08-13 01:33:28)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Macy

Guest

Re: Algebra headache

Geezzz... you guys make me impressed by:

1. Fast response
2. Your different approach
3. Your enthousiasm

big Hug to all

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra headache

Hi Macy;

Thanks from all 3 of us.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juriguen

Member

Registered: 2009-07-05

Posts: 59

Re: Algebra headache

Yes, thanks Macy!

I like the other approaches You can always learn many new things in this forum, even if you got the correct answer!

Jose

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“Make everything as simple as possible, but not simpler.” -- Albert Einstein

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Algebra headache

Everyone here has knowledge, so they are all my friends.

Last edited by bobbym (2009-08-13 20:05:08)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Re: Algebra headache

Nothing like beating a dead horse, but here we go...

So the last digit is either 3 or 8, since all of the factors are even the last digit must be an 8.

Last edited by Fruityloop (2009-08-27 09:56:37)