[Guest]

Inspired by IdoJava's equations, I wondered whether we can find some minimal solutions for similar problems. So,

For each number [1-9], can you find the minimal combination (ie uses the figure the fewest number of times) which results in a value of 10?

For example,

1+1+1+1+1+1+1+1+1+1 [10] would be a soultion for 1, but

(1+1+1+1+1) × (1+1) [7] would be better, and obviously there are even better solutions

For each number n, you may only use 'n' or '-n' and the four basis operators (+, -, ×, ÷). Oh, and parentheses wher necessary, but no roots or powers (ie. you must use 2×2 rather than 2²)

Some are definitely easy, but some I am struggling to see a short and sweet answer....

I don't have definite answers to these, but had these thoughts while making this:

1. We noted with IdoJava that 6 seemed to be an 'easy' choice for the result, as 6=3!. Is 10 an 'easy' or 'hard' choice and are there types of numbers which are always 'easy'?

2. How much better are the solutions if you include the 'raise to power' operator, ^, and/or the factorial operator, !?

note: there may have been similar questions asked before, so apologies if it has already been done: I did try to search!

[Guest]

((1+1+1)^(1+1))+1

[HoustonHokie]

5+5=10 (used two times)

And I'll try to actually use my brain for some of the other 8 digits!

[TwoaDay]

(9/9)+9

3 times

2x2x2+2

4 times

Edited August 13, 2008 by TwoaDay

[Guest]

(2*2*2)+2=10 [4]

3*3 + 3/3 = 10 [4]

4+4 + (4+4)/4 = 10 [5]

5+5=10 [2]

6+6 - (6+6)/6 = 10 [5]

[Guest]

((1+1+1)^(1+1))+1

I think the OP said no powers.

(1+1+1)*(1+1+1) + 1 = 10 [7]

[Guest]

7 + 7/7 + (7+7)/7 = 10 [6]

8 + (8+8)/8 = 10 [4]

[Prof. Templeton]

Do you have to put an operation between every digit? Otherwise you would need at most 4 digits.

[44-4]/4=10

[55-5]/5=10

and so on

[HoustonHokie]

7+(7+7+7)/7 => 5

[Guest]

7+(7+7+7)/7 => 5

oh, of course. silly me.

[Guest]

I'm thinking it's not possible to beat 7 using ones. Unless you don't need an operation between each numeral, in which case

11-1=>3

[HoustonHokie]

the only other one I found that could be improved by adding ^ or ! was 4:

4!/4 + 4 = 10 [3]

A solution for 1 has already been provided.

I don't know if we could say this was easy or hard. Certainly seems harder than 6, but it's also larger than 6 and so its difficult to get the small numbers all the way there. Factorial numbers like 6 & 24 may be easier than other numbers if the factorial is allowed because you can get a long way in one step for a particular digit (and that might translate to other digits as sums and products, i.e. (2+2)! = 24). Primes are probably the most difficult because you can't reach them by direct multiplication or addition - always a multi-step process. I guess we could try 7, 8, & 9 to start and compare? My gut is that 7 would be really hard (prime) and 8 & 9 would be comparable to 10. Who knows what happens with larger 2 and 3 digit numbers?

[unreality]

(1+1)*(1+(1+1)(1+1)) = 10 [7]

(1+1)(1+1+1+1+1) = 10 [7]

(1+1+1)(1+1+1)+1 = 10 [7]

(2+2)*2 + 2 = 10 [4]

2*2*2 + 2 = 10 [4]

3/3 + 3*3 = 10 [4]

4*((4+4)/4) + ((4+4)/4) = 10 [7]

5+5 = 10 [2]

6+6 - ((6+6)/6) = 10 [5]

7+7 - ((7+7+7+7)/7) = 10 [7]

8+ 8/8 + 8/8 = 10 [5]

9+ 9/9 = 10 [3]

10 = 10 [1]

there are probably smaller answers to some of those

[Guest]

Looks like Cherry Lane has it from HoustonHokie 4-2, unless there are some better ones... I find it interesting that 4 can't be bettered - I always think of 4 as quite a cooperative number!

(1+1+1)*(1+1+1) + 1	 Cherry Lane [7] <- I assumed there would be better solutions than my [7], but maybe not!

2x2x2+2				 TwoaDay [4]

3x3 + 3÷3			   Cherry Lane [4]

4+4 + (4+4)/4		   Cherry Lane [5]

5+5					 HoustonHokie [2] <- I don't think you'll beat this!

6+6 - (6+6)/6		   Cherry Lane [5]

7+(7+7+7)/7			 HoustonHokie [5]

8 + (8+8)/8			 Cherry Lane [4]

9+(9÷9)				 TwoaDay [3]

My other random thought is that there is almost some symetry there, though that might be because of the choice of '10' in a base 10 system.

Do you have to put an operation between every digit? Otherwise you would need at most 4 digits.

[44-4]/4=10

[55-5]/5=10

and so on

Yep. As you say, it would be fairly trivial otherwise!

[Guest]

(9/9)+9