# Tag Info

### Visiting every digit

Proof that Daniel Mathias' solution is optimal:
• 3,071
Accepted

### Visiting every digit

The permutation requires a total of This is a slight alteration (a partial reversal) of Culver Kwan's excellent find:
• 15.4k

### Largest sequence of adjacent numbers less than 11 such that adjacent number divides the other

We work backwards from our constraints, starting from the most restrictive numbers. 7 can only go next to 1: 7 1 Next 5 and 9 only have 2 options, either 1 or 10 and 3, respectively. So one of these ...
• 91

### Largest sequence of adjacent numbers less than 11 such that adjacent number divides the other

The "divisor" graph $G$ has node set $\{1,2,\dots,10\}$ and an edge $(i,j)$ if $i$ divides $j$ or $j$ divides $i$. Paths of length $9$ are easy to find. To show that $10$ is impossible, we ...
• 14.3k

### Visiting every digit

I constructed With I think it is optimal?
• 6,229

• 824

### 2,3,6,7 to get 10

Just using the four basic operations:
• 3,071

### Largest sequence of adjacent numbers less than 11 such that adjacent number divides the other

I can get nine of the ten numbers: Explanation: I'm not aware of the exact mental process of how I found a solution, but this is the entire jotting of how it emerged: My proof of the maximum:
• 14.6k

### A 3 digit perfect square and its reverse are both perfect squares. What is the number?

Let $[a_r,\dotsc,a_0]$ denote $\sum_{i=0}^r a_i10^i$ with $0\leq a_i\leq 9$. Let $0\leq n \leq 999$ be a perfect square, say $n=m^2$, where $m=[a,b]$. Note that $m\leq 31$, so that $0\leq a\leq 3$. ...
• 31
Accepted

### 2,3,6,7 to get 10

Using basic operations and a two digit number:
• 13.2k
1 vote

### 2,3,6,7 to get 10

You didn't specify whether exponents and factorials are allowed so I'm assuming they are. I found 2 answers: 1: 2:
• 438

Only top scored, non community-wiki answers of a minimum length are eligible