Thanks to Yale from Hong Kong for suggesting this puzzle! The problem went viral after a video from Carl Ho (in Chinese): https://youtu.be/6acBMe0Ugy8.
Here is the problem: using three 3s, can you make the numbers from 0 to 10 using mathematical operations?
3 3 3 = 0
3 3 3 = 1
3 3 3 = 2
3 3 3 = 3
3 3 3 = 4
3 3 3 = 5
3 3 3 = 6
3 3 3 = 7
3 3 3 = 8
3 3 3 = 9
3 3 3 = 10
There are many ways to solve this, and I will present some of answers.
The hardest problem is 3 3 3 = 10. My challenge, which is a very hard challenge, is to solve this in two ways using only the following symbols:
+ ! ()
There is a way to solve 3 3 3 = 10 using only those symbols, although the notation is lesser known.
Can you figure it out? Watch the video for a solution.
Can You Solve The Three 3s Challenge?
Or keep reading. "All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To The 3s Challenge
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Here are some ways to get 0 to 9. (You may have found other solutions not listed here.)
3! – 3 – 3 = 0
(3! – 3)/3 = 1
33 – 3 = 1
√3 × √3 / 3 = 1
3 – 3/3 = 2
(3 + 3)/3 = 2
3 + 3 – 3 = 3
3 × 3/3 = 3
3 + 3/3 = 4
3! – 3!/3 = 4
3! – 3/3 = 5
3!/3 + 3 = 5
3 × 3 – 3 = 6
3! × 3/3 = 6
3 << (3/3) = 3×21 = 6 (<< is the left arithmetic shift)
3! + 3/3 = 7
3! + (3 – 3)! = 7
3/.3 – 3 = 7
3!/3 + 3! = 8
3 + 3 + 3 = 9
33/3 = 9
How can we make 10? Here are some of the creative ways I found.
In Carl Ho’s video comments, someone suggested:
arctan(√3)/(3 + 3) = 10 (degrees)
The idea is that arctan(sqrt(3)) is 60 degrees, so you then divide by 6 to get 10. The answer is technically in degrees, but for this puzzle we may allow this creative answer.
This is an answer from Cut The Knot:
floor(√3 + √3)×3 = 10
Yale suggested three methods to me as well.
√(√(3 × 3)/3%) = 10
log√√3(3) + 3! = 10
3 × 3 + (d/dx (3))! = 10
Now for the challenge: how can we solve it using only +, ! and ( )?
Here are the solutions:
3! + !3 + !3 = 6 + 2 + 2 = 10
!3(!3 + 3) = 2(2 + 3) = 10
How is !3 = 2? This requires a bit of explanation.
The notation n! denotes the factorial of n, which is the total number of permutations of n.
By contrast, the notation !n denotes the subfactorial of n, which is the number of derangements of n. A derangement is a permutations in which no item is in its natural order.
For example,
!1 = 0 because {1} has no derangements
!2 = 1 because {1, 2} has the single derangement {2, 1}
!3 = 2 because {1, 2, 3} has two derangements: {2, 3, 1} and {3, 1, 2}
While this might be new to you, it is accepted enough notation that even WolframAlpha will evaluate. You can see:
!3(!3 + 3) evaluates to 10
It’s a sneaky way to use the exclamation point, but in this puzzle the subfactorial is very useful! You’ll also find extra solutions to the other equations using !3 = 2.
Another bonus answer
I received this by email:
!(!3 + !3) + !(!3) = 10
This uses that !4 = 9 and !2 = 1 which is really clever!
(Note: you can get the number of derangements by the formula floor(n!/e + 1/2) for n ≥ 1. So you will note the number of derangements, divided by the number of permutations n!, is roughly 1/e ≈ 37%. It’s kind of interesting the ratio of derangements to permutations is roughly constant even as n increases!)
Sources
Carl Ho’s video (Chinese)
https://youtu.be/6acBMe0Ugy8
Cut The Knot Representation of numbers with three 3’s
http://www.cut-the-knot.org/arithmetic/funny/3_3.shtml
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Kyle
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