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.
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4 Answers
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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
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$\begingroup$ im not sure if we can add ^0, ^2 or ^3, isn't that adding numbers? $\endgroup$– KepotxCommented Mar 20, 2018 at 14:25
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$\begingroup$ @Kepotx Not really a number, it is a mathematical operator but we write it with numbers. $\endgroup$– SaeïdrylCommented Mar 20, 2018 at 14:27
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$\begingroup$ @Kepotx Ok: 10-(log(10*10*10)) = 7 $\endgroup$ Commented Mar 20, 2018 at 14:28
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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
answered Mar 20, 2018 at 14:02
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$\begingroup$ Hahahahaha, this is so clever :P $\endgroup$ Commented Mar 20, 2018 at 14:05
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1$\begingroup$ For anyone wondering why these types of puzzles need to be well-specified, this answer is why. $\endgroup$– F1KrazyCommented Mar 20, 2018 at 14:09
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1$\begingroup$ 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 $\endgroup$ Commented Mar 20, 2018 at 14:10
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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
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$\begingroup$ no, cant separate the all 10 numbers $\endgroup$– PeggyCommented Mar 20, 2018 at 14:36
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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"
+-*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) $\endgroup$