The Puzzle:
How can I get the answer 24 by only using the numbers 8,8,3,3.
You can use add, subtract, multiply, divide, and parentheses.
Bonus rules: also allowed are logarithms, factorials and roots
(Puzzle supplied by "Steve123")
You can use add, subtract, multiply, divide, and parentheses.
Bonus rules: also allowed are logarithms, factorials and roots
(Puzzle supplied by "Steve123")
Our Solution:
1) Supplied by "mathsyperson":
8/(3-(8/3))
= 8/(1/3)
= 24
2) Supplied by "puzzler09" (using bonus rules):
((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24
3) Supplied by "Mark" (using bonus rules):
(3!/)*8
4) Supplied by "Daryl S" (using bonus rules):
(8-3)!/(8-3)
( × )!
( + )!
√(8×8×3×3)
8+(8×(3!/3))
((√(8+8) × (3/3))!
√(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!
5) Supplied by "Sunil Prajapati" (using bonus rules):
√(8×8)×√(3×3) which is a variation of √(8×8×3×3) by Daryl S
8/(3-(8/3))
= 8/(1/3)
= 24
2) Supplied by "puzzler09" (using bonus rules):
((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24
3) Supplied by "Mark" (using bonus rules):
(3!/)*8
4) Supplied by "Daryl S" (using bonus rules):
(8-3)!/(8-3)
( × )!
( + )!
√(8×8×3×3)
8+(8×(3!/3))
((√(8+8) × (3/3))!
√(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!
5) Supplied by "Sunil Prajapati" (using bonus rules):
√(8×8)×√(3×3) which is a variation of √(8×8×3×3) by Daryl S
No comments:
Post a Comment