Math Is Fun Forum

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,395

Proofs required!

I urgently require two proofs.

1. That 2^4(5^m*n) keeps giving last places which repaet themselves perfectly in order. Like, 2^4 ends in 6, 2^20 ends in 76 and soon.

2. That the only way of getting such numbers for those ending in 6 is as above.

Matter most urgent, top priority may please be accorded to this.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Proofs required!

2^n follows a pattern in the last digit of (for natural numbers): (starting at 2^1) 2,4,8,6

so:
2^{4n-3} ends in 2
2^{4n-2} ends in 4
2^{4n-1} ends in 8
2^{4n} ends in 6

n ∈ N

so it can be reduced to: 4(5^m*n) is divisable by 4, which is then simple, since its 4 times 5^m*n its always divisable by 4 so 2^4(5^m*n) always ends in 6 (altleast for all m,n ∈ I)

Now supposing you need a proof for my pattern, that is perhaps a different matter, i hope you did mean m and n to be integers so that the power of 2 is always 1 or more

You can see why that pattern exists, you start at 2, multiply by 2, you get 4, multiply by 2 you get 8, multiply by 2 you get 16 (6), 6×2 is 12 (2), and back to the beginning, since the other digits don't effect the last digit that pattern emerges, im not sure how you would give a formal proof mind.

Last edited by luca-deltodesco (2007-12-08 22:07:32)

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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Proofs required!

@Ganesh: Why do you need m*n? Just one variable will do! n, where n is a non-negative integer.

@Luca: The question is not about last digit. It's about last digits (plural).

etc.

Last edited by JaneFairfax (2007-12-08 23:21:12)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Proofs required!

m is a constant that determines how many last digits there are (there are m+1), and n is a variable.

Your LaTeX should read:

If you wanted to get even more general, you could even put a "+c" after each n in that array, but that's a simple corollary of ganesh's proposition anyway, so we don't need to prove that bit.

Edit: This topic is related at one point, although the subject keeps changing.

Edit2: Just realised I should clarify, adding the "+c" will mean that the numbers won't necessarily have the last digits quoted above, but they will be the same for any n.

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Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,395

Re: Proofs required!

Thanks luca-deltodesco, JaneFairfax and mathsyperson!
I shall study the proofs 100% and check for any possible loopholes, which at present I don't think exist!

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.