# find out the missing expressions [closed]

10 10 10 10 = 1
10 10 10 10 = 2
10 10 10 10 = 3
10 10 10 10 = 4
10 10 10 10 = 5
10 10 10 10 = 6
10 10 10 10 = 7
10 10 10 10 = 8
10 10 10 10 = 9
10 10 10 10 = 10

You can only add expressions, you cannot change any numbers.

• What are the allowed mathematical operators? – Saeïdryl Mar 20 '18 at 13:53
• Hi Peggy, and welcome! When you make this kind of problems, you need to say what we can use. Is it just the symbols + - * and /, or maybe something more? (If we can use anything, then it's not a very interesting problem anymore. We already know how to solve all of those) – Bass Mar 20 '18 at 14:02
• all symbols, including brackets and roots – Peggy Mar 20 '18 at 14:38
• @Peggy Could you please give us a list of symbols? As Bass pointed, if you allow all symbols the solution is easy. – Saeïdryl Mar 20 '18 at 14:59
• I did some of them; – Peggy Mar 20 '18 at 16:28

Complete answer with "simple" operators (+, -, *, /, ⁰, ², ³, !, log):

(10 + 10) / (10 + 10) = 1
(10 / 10) + (10 / 10) = 2
(10 + 10 + 10) / 10 = 3
((10 / 10) + (10 / 10))² = 4
Another one : log(10) + log(10) + log(10) + log(10) = 4
(10 * 10) / (10 + 10) = 5
((10 + 10 + 10) / 10)! = 6 (thanks to Kepotx)
10 - 10⁰ - 10⁰ - 10⁰ = 7 (thanks to Stefano Lonati)
Another one : 10 - log(10 * 10 * 10) = 7 (thanks to Stefano Lonati)
((10 / 10) + (10 / 10))³ = 8
Another one : 10 * log(10) - log(10) - log(10) = 8 (thanks to Stefano Lonati)
((10 * 10) - 10) / 10 = 9
10 + (10 - 10) * 10 = 10

• (10 * 10) - 10 - 10 = 80, not 8 – Kepotx Mar 20 '18 at 14:10
• im not sure if we can add ^0, ^2 or ^3, isn't that adding numbers? – Kepotx Mar 20 '18 at 14:25
• @Kepotx Not really a number, it is a mathematical operator but we write it with numbers. – Saeïdryl Mar 20 '18 at 14:27
• @Kepotx Ok: 10-(log(10*10*10)) = 7 – Stefano Lonati Mar 20 '18 at 14:28
• Hi Anyone can explain the ; – Peggy Mar 20 '18 at 23:59

Thinking a little outside of the box, here's my answer:

10 + 10 + 10 + 10 != 1
10 + 10 + 10 + 10 != 2
10 + 10 + 10 + 10 != 3
10 + 10 + 10 + 10 != 4
10 + 10 + 10 + 10 != 5
10 + 10 + 10 + 10 != 6
10 + 10 + 10 + 10 != 7
10 + 10 + 10 + 10 != 8
10 + 10 + 10 + 10 != 9
10 + 10 + 10 + 10 != 10

where the operators I used were:

+ - the addition operator
! - logical NOT

This answer can be extended to any $n$ for {$n \in \mathbb N, \mathbb Z, \mathbb R, \mathbb Q\}$ with only one exception for

n = 40

• Hahahahaha, this is so clever :P – ABcDexter Mar 20 '18 at 14:05
• For anyone wondering why these types of puzzles need to be well-specified, this answer is why. – F1Krazy Mar 20 '18 at 14:09
• To be quite honest I expected this answer to be down voted, but didn't mind since it showed the OP the need to be specific – James Webster Mar 20 '18 at 14:10

Is it possible separate the number 10 in 1 and 0?

1*0+1*0+1*0+1+0 = 1
1*0+1*0+1+0+1+0 = 2
1*0+1+0+1+0+1+0 = 3
1+0+1+0+1+0+1+0 = 4
10/(1+0+1+0)+ 1*0 = 5
10/(1+0+1+0)+ 1 + 0 = 6
10-1+0-1+0-1+0 = 7
10-1+0-1+0+1*0 = 8
10-1+0+1*0+1*0 = 9
10-1*0-1*0-1*0 = 10

Or

(10 + 10) / (10 + 10) = 1
(10 / 10) + (10 / 10) = 2
(10 + 10 + 10) / 10 = 3
log(10*10) + log(10*10) = 4
(10 * 10) / (10 + 10) = 5
((10 + 10 + 10) / 10)! = 6
10 - log(10*10*10) = 7
10 * log(10) - log(10) - log(10) = 8
((10 * 10) - 10) / 10 = 9
10 + (10 - 10) * 10 = 10

• no, cant separate the all 10 numbers – Peggy Mar 20 '18 at 14:36

A simple general solution using Log, +,-,*,(),^

$$10*(10 - 10) + \log{(10^1)} = 1$$ $$10*(10 - 10) + \log{(10^2)} = 2$$ $$10*(10 - 10) + \log{(10^3)} = 3$$ $$10*(10 - 10) + \log{(10^4)} = 4$$ $$10*(10 - 10) + \log{(10^5)} = 5$$ $$10*(10 - 10) + \log{(10^6)} = 6$$ $$10*(10 - 10) + \log{(10^7)} = 7$$ $$10*(10 - 10) + \log{(10^8)} = 8$$ $$10*(10 - 10) + \log{(10^9)} = 9$$ $$10*(10 - 10) + \log{(10^{10})} = 10$$

This simple approach can generate any "n"