119
votes
114
votes
Accepted
93
votes
Accepted
78
votes
75
votes
74
votes
64
votes
64
votes
Accepted
62
votes
Accepted
60
votes
Accepted
58
votes
57
votes
Accepted
Create all numbers from 0-100 only using all of 1,2,3,4 and 5
Also, you can use any operation.
Ok then.
$\begin{array}{c|c}
0 & \log_{\frac1 2} \left( \log_{4!!-3} 5 \right) \\
1 & \log_{\frac1 2} \left( \log_{4!!-3} \sqrt 5 \right) \\
2 & \log_{\...
- 763
54
votes
Accepted
Making 103 from 4 zeroes
As
rand al’thor points out
in the solution built upon here,
there must be a way to formulate a
$\small 3$ with only
two $\small 0 \kern1mu$s.
How promising that...
&...
- 21.9k
53
votes
53
votes
52
votes
Accepted
A clock for 2017
With the digits in order:
$$
\begin{align}
1 &= 2 + 0 - 1 ^ 7 \\
2 &= 2 + 0 \times 1 \times 7 \\
3 &= 2 + 0 + 1 ^ 7 \\
4 &= -2 - 0 - 1 + 7 \\
5 &= 2 \times (0 - 1) + 7 \\
6 &= ...
- 19k
52
votes
Accepted
Make 5 5 5 5 = 19
Here's one way I found:
Or, using just the characters explicitly allowed in the question:
- 14.5k
48
votes
A clock for 2017
I tried to make a digital clock.
$0 = (7 + 1 + 2) \times 0$
$1 = (2 + 7 + 1) ^ 0$
$2 = (7 + 1) \times 0 + 2$
$3 = 7 \times 0 + 2 + 1$
$4 = 2 \times 7 - 10$
$5 = 7 - 2 + 1 \times 0$
$6 = 7 - 1 + 2 \...
- 18.4k
48
votes
Accepted
Doubling/tripling puzzle: make 1 from 1536 in as few steps as possible
As Jo has already shown, this can be accomplished in
To help visualize this problem, we can imagine:
Proving minimality:
- 7,137
48
votes
Combine 1,3,3,7 to get 10
As quite standard in this kind of hard number puzzle, we can:
Another example of this form being the only solution is:
Use 1, 2, 3, 8 to make 28
with the unique (up to commutation) solution being:
- 6,059
43
votes
43
votes
How to get 5 from 0,0,0 and 1?
First off, latest edit - just for fun, how to get 5 from just 0 and 1:
Before rule change posted:
With the changed rules:
And while we're at it, here's $0$ to $28$:
And here's how to get 5 from ...
- 9,431
43
votes
42
votes
41
votes
Create all numbers from 0-100 only using all of 1,2,3,4 and 5
$\begin{align}
0 & = (1 + 2 - 3) \times (4 + 5) \\
1 & = 1 + (2+3-5) \times 4 \\
2 & = 2 + (1+3-4) \times 5 \\
3 & = 1 -2+3-4+5 \\
4 & = 1 \times (2+3-5) + 4 \\
5 & = 1-2-3+4+...
- 18.4k
40
votes
How to get 32 by using +1 , +1 , ×3 , ×3 , ÷2 , ÷2, ^2, ^2?
I like Glorfindel's answer as it takes you through an example solution, and I recommend his answer. Just to show some underlying beauty and complexity of this puzzle, I created a diagram (directed ...
- 2,296
40
votes
How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed
Well, here is how I would (and did) solve it:
First step: Without the « five + » constraint
Second step: Getting rid of some « + »
A solution we found:
Extra note: finding all solutions
- 5,096
39
votes
38
votes
Accepted
38
votes
Accepted
How should I approach using two 8s and two 3s to make the number 24?
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the ...
- 1,200
Only top scored, non community-wiki answers of a minimum length are eligible