[Guest]

a wise man wanted to develop a curreny with certain increments with the following properties.

1) no more than a single unit of each increment is nessicary to reach the total desired.

2) no consecutive increments are nessicary to reach the total desired.

for example, the following increments would be undisireable.

1 2 4 8

because to get 3, you need 1 and 2, and to get 7 you need 4 2 and 1.

if only addition is allowed and the wise man wanted every value between 1 and 100 represented with the fewest increments, what increments should be used?

[Guest]

The numbers that I think you need are

1, 2, 3, 4, 5, 10, 20, 30, 40, 50

MY REASONING:

I started off by working my trough the numbers 1-10.

To make 1 you would need to use the unit 1.

To make 2 you would need to use the unit 2 to follow the rules.

To make 3 you would need to use the unit 3 to follow the rules.

To make 4 you could use 1+3

To make 5 you need to use the units 1+4 to follow the rules.

To make 6 you need to use 2+4

To make 7 you need to use 2+5

To make 8 you need to use 3+5

To make 9 you need to use 1+3+5

Then all I needed to do was make each of the units I used into tens number so 1=10, 2=20 ......

This assured me that I could make the numbers 1-9 and add them onto 10/20/30/40/50/60/70/80/90 to make whatever number required and follow the puzzles rules

[k-man]

follow the fibonacci sequence: 1,2,3,5,8,13,21,34,55,89

[Guest]

k man got it! i figured this would be solved quickly.

[superprismatic]

k man got it! i figured this would be solved quickly.

What's wrong with Riddle_Lover's answer? It looks good to me, but perhaps I don't understand the rules.

[fabpig]

What's wrong with Riddle_Lover's answer? It looks good to me, but perhaps I don't understand the rules.

(eg) to make 17 you'd need 10 + 5 + 2, and '10 and 5 ' are "consecutive increments". That's my reading of it,anyway. But at first I thought it meant consecutive as in 1 and 2 or 3 and 4, too

Edited May 26, 2011 by fabpig

[superprismatic]

(eg) to make 17 you'd need 10 + 5 + 2, and '10 and 5 ' are "consecutive increments". That's my reading of it,anyway. But at first I thought it meant consecutive as in 1 and 2 or 3 and 4, too

Oh! Thanks, fabpig, I would never have thought of that.

[plainglazed]

working off of the above - 1, 2, 3, 4, 5, 10, 17, 27, 44, 71

[Guest]

Would there be any other kinds? I was think that MAYBE you could use

Primes? 1,2,3,5,7,11,13,17,19,23,29...

I probably don't think I'm right.

[Guest]

working off of the above - 1, 2, 3, 4, 5, 10, 17, 27, 44, 71

This will also work, but k-man's solution is general and can be applied for any number.

Edited May 27, 2011 by Mukul Verma

[Guest]

follow the fibonacci sequence: 1,2,3,5,8,13,21,34,55,89

Having found a novel solution, can we prove it and generalise it?

[k-man]

Having found a novel solution, can we prove it and generalise it?

Not sure what you mean by generalizing it, but a proof is pretty straight forward.

With the numbers 1 and 2 we can produce numbers 1 and 2. This is our initial set.

Given:

We have a set of ordered distinct natural numbers. Size of the set is at least 2.

Let's call the largest 2 numbers in the set X and Y (X<Y).

By adding the numbers from the set we can generate any natural number up to and including X+Y-1 with the constraints that

1) each number from the set can be used only once

2) two numbers that appear consecutive in the set cannot be used together

Prove that by adding the new member Z=X+Y to the set we will be able to generate all natural numbers up to an including Y+Z-1 using the same rules above.

Proof:

Y is the only "neighbor" of Z, so Z can be used with any other valid combination of numbers that doesn't include Y. Since the original set allows us to generate all natural numbers up to and including X+Y-1 it allows us to generate all natural numbers up to and including Y-1 without using Y. So, if we add Z to all those numbers we will get the range [Z+1,Z+Y-1] still without using Y and without violating the rules. Since Z=X+Y, we can generate

1) all numbers in the range [1,Z-1] without using Z (based on the original set)

2) number Z

3) and all numbers in the range [Z+1,Z+Y-1] without violating the rules.

Proof complete.