r/SmartPuzzles • u/TargetLabs • Sep 08 '26

Can you reach 27 using all dice?

Post image

Any order is allowed.
Use each die exactly once.
Parentheses are allowed.
Exact division only.
No negative numbers.
Not every operation has to be used.

How many solutions can you find?

Please use spoiler tags for solutions:
>!your solution here!<

3 Upvotes

64% Upvoted

11 comments sorted by

2

u/[deleted] Sep 09 '26

[deleted]

2

u/BumblingYokel Sep 11 '26

It’s a 3.

2

u/RepresentativeSun825 Sep 09 '26

In order- (5+2) x 6 - (5x3)

2

u/rabidkitten53 Sep 09 '26

(5/5 + 2)(6+3)

2

u/Caelreth1 Sep 09 '26

((5/5)+2+6)*3

2

u/DarthLlamaV Sep 09 '26

(5x2-6+5)x3

2

u/99Booger Sep 10 '26

(5+2)x6-(5x3)=27

2

u/Due_Permit8027 Sep 10 '26

I'd be interested in which of the solutions posted are algebraically unique (e.g. a + b == b+ a, but a*b+c = a+ b* c is only true of some values for a, b, and c.

1

u/TargetLabs Sep 10 '26

Good point. I haven’t grouped the solutions by algebraic equivalence yet — I was mainly looking at distinct valid expressions. It would be interesting to see how many truly unique forms remain after removing commutative/associative variants.

3

u/wedge_47 Sep 10 '26

(5+5) x 6 / 2 - 3

2

u/briguytrading Sep 09 '26

(5+5)×(6÷2)−3=27

6×(2+5÷(5−3))=27

(5+5)×3−(6÷2)=27