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# only allowed 1

### #1

Posted 26 August 2012 - 04:52 AM

further more the only number you are allowed to use is 1. for example, the number 15 could be

(1+1+1)*((1+1)*(1+1)+1)

express 137 with the fewest number of 1's in a single expression.

### #2

Posted 26 August 2012 - 08:56 AM

### #3

Posted 26 August 2012 - 05:39 PM

### #4

Posted 26 August 2012 - 07:48 PM

### #5

Posted 27 August 2012 - 03:10 PM

^{ Spoiler for thought I had 14 ...but I forgot about the two at the end ((1+1+1)(1+1+1)(1+1+1)(1+1+1+1+1))+1+1 }

### #6

Posted 28 August 2012 - 06:38 PM

### #7

Posted 29 August 2012 - 03:05 AM

EXTENSION PROBLEM. What is the smallest number that requires at least 16 ones--and what is the largest number that can be expressed with no more than 16 ones?

**Edited by TheChad08, 29 August 2012 - 03:07 AM.**

### #8

Posted 29 August 2012 - 04:45 AM

EXTENSION PROBLEM. What is the smallest number that requires at least 16 ones--and what is the largest number that can be expressed with no more than 16 ones?

Let's restate to avoid misinterpretation (given the conditions set forth in OP about 1's +, x and parentheses):

1. What is the smallest number that cannot be expressed using fewer than 16 1's?

2. What is the largest number that can be expressed using 16 1's or fewer?

Since we can add any leftover 1's if fewer than 16 are used,

**precisely**16 1's applies in the second case.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #9

Posted 29 August 2012 - 05:22 AM

Let's restate to avoid misinterpretation (given the conditions set forth in OP about 1's +, x and parentheses):

1. What is the smallest number that cannot be expressed using fewer than 16 1's?

Well, that requires a LOT of work.

We would need to know what numbers are possible for the other combinations.

Sounds like more work than I want to do.

Someone will write a code with all possible combinations of numbers to get it.

I see 32 768 combinations of addition and multiplication between the 15 ones. This doesn't include the brackets.

Lots of these will be mirror images though and can be discounted.

**Edited by TheChad08, 29 August 2012 - 05:23 AM.**

### #10

Posted 29 August 2012 - 04:46 PM

Well, that requires a LOT of work.

We would need to know what numbers are possible for the other combinations.

Sounds like more work than I want to do.

Someone will write a code with all possible combinations of numbers to get it.

I see 32 768 combinations of addition and multiplication between the 15 ones. This doesn't include the brackets.

Lots of these will be mirror images though and can be discounted.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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