Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,089

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

Offline

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

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)*

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

@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 digit**s** (plural).

etc.

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

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Why did the vector cross the road?

It wanted to be normal.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,089

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!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

Offline

Pages: **1**