Integers = sums of 2s and 3s.
 

I stumbled upon this one a while ago, as I posted elsewhere:

Initially, I thought it was only the primes that could be decomposed into 2s and 3s. Then, I realized I could generalize this to all integers greater than 3.

Here are some examples:
47 = 2(19) + 3(3)
79 = 3(17) + 2(14)
106 = 3(20) + 2(23)
189 = 2(60) + 3(23)

Etc.

Proving it might be an easy task. Tips?

Also: Congrats to me on a Fermat prime post. (257) --> 2^(2^3)+1.

[QUOTE=3.14159;222093]I stumbled upon this one a while ago, as I posted elsewhere:

Initially, I thought it was only the primes that could be decomposed into 2s and 3s. Then, I realized I could generalize this to all integers greater than 3.

Here are some examples:
47 = 2(19) + 3(3)
79 = 3(17) + 2(14)
106 = 3(20) + 2(23)
189 = 2(60) + 3(23)

Etc.

Proving it might be an easy task. Tips?

Also: Congrats to me on a Fermat prime post. (257) --> 2^(2^3)+1.[/QUOTE]

Generalize:

Let m and n be elements of Z+ such that (m,n) = 1.

Ask yourself: What is the smallest integer M that is [b]not[/b]
representable as mx + ny???

This is a very well known problem. Google is your friend.

Searchbait: Sylvester, Frobenius, "happy meal"

[QUOTE]Ask yourself: What is the smallest integer M that is [B]not[/B]
representable as mx + ny???[/QUOTE]

For x = 2; y = 3; I doubt that there are any counterexamples.

The only counterexample I've found is 1.

A probable counterexample is 4: 2+2, since there are no 3s in that decomposition.

It's looking like the counterexamples are powers of 2:
Let's test a few:
8 = 3(2) + 2
16 = 3(4) + 2(2).

Powers of 3:
9: 2(3) + 3
27: 3(5) + 2(6).

The only counterexamples I've found are {1, 4}.

However, when using larger integer pairs, it fails horrendously:
Let's use 6 and 7:
8 can't be represented as 6a + 7b
4 and 5:
10: 5+5
11: Cannot be expressed as 4a + 5b.

The only pair for which it works so well is 2 and 3 (Also 1 and 2, the best pair of all.)

[QUOTE=3.14159;222097]However, when using larger integer pairs, it fails horrendously:
Let's use 6 and 7:
8 can't be represented as 6a + 7b[/QUOTE]

29 can't be represented, either. What about larger numbers?

[QUOTE=3.14159;222093]I stumbled upon this one a while ago, as I posted elsewhere:

Initially, I thought it was only the primes that could be decomposed into 2s and 3s. Then, I realized I could generalize this to all integers greater than 3.
[/QUOTE]

If it's even, it's a multiple of two; if you feel you need to have some threes in the mix, n-6 is a multiple of two plus 2*3

If it's greater than three and odd, then n-3 is a multiple of two.

This is not hard problem.

[QUOTE]If it's even, it's a multiple of two; if you feel you need to have some threes in the mix, n-6 is a multiple of two plus 2*3

If it's greater than three and odd, then n-3 is a multiple of two.

This is not hard problem.
[/QUOTE]

I wasn't stating that it was difficult. I was curious to know whether or not it was proven. It seemed interesting, is all.

[QUOTE]29 can't be represented, either. What about larger numbers?[/QUOTE]

108: 7(12) + 6(4)
109: 7(7) + 6(10)
110: 6(2) + 7(14)
111: 6(15) + 7(3)
112: 6(14) + 7(4)
113: 7(5) + 6(13)

Etc. What's the final counterexample here?

[QUOTE=3.14159;222102]What's the final counterexample here?[/QUOTE]

29.

[QUOTE=3.14159;222101]I wasn't stating that it was difficult. I was curious to know whether or not it was proven. .[/QUOTE]


But clearly not sufficiently curious to have done some work yourself
by using Google. You were given hints.

[QUOTE]But clearly not sufficiently curious to have done some work yourself
by using Google. You were given hints.[/QUOTE]

I actually did find it. Stop making false assumptions:
[URL="http://en.wikipedia.org/wiki/Coin_problem"]Here it is.[/URL]

[URL="http://mathworld.wolfram.com/CoinProblem.html"]A better source.[/URL]

[QUOTE=3.14159;222123]I actually did find it. Stop making false assumptions:
[URL="http://en.wikipedia.org/wiki/Coin_problem"]Here it is.[/URL]

[URL="http://mathworld.wolfram.com/CoinProblem.html"]A better source.[/URL][/QUOTE]

Then why did you have to ask if it were proven?

[QUOTE]Then why did you have to ask if it were proven?[/QUOTE]

I previously didn't know what it was. I merely followed your tip + CRG's "searchbait".