### Project Euler Problem 32 Statement

We shall say that an `n`-digit number is pandigital if it makes use of all the digits 1 to `n` exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

### Solution

This is a closed–ended recreational problem, so not much thought was given to a clever solution, but here's a table of results I created before trying to solve the problem:

Total Pandigital Digits | Digits Factor 1 | Digits Factor 2 | Digits Product | Maximum Factor 1 | Maximum Factor 2 | Solutions | Sum of products |
---|---|---|---|---|---|---|---|

4 | 1 | 1 | 2 | 3 | 4 | 3×4=12 | 12 |

5 | 1 | 2 | 2 | 4 | 13 | 4×13=52 | 52 |

6 | 1 | 2 | 3 | 3 | 54 | 3×54=162 | 162 |

7 | None | 0 | |||||

8 | 1 | 3 | 4 | 6 | 582 | 3×582=1746 6×453=2718 | |

8 | 2 | 2 | 4 | 58 | 64 | 24×57=1368 34×52=1768 37×58=2146 58×64=3712 | 13458 |

9 | 1 | 4 | 4 | 4 | 1963 | 4×1738=6952 4×1963=7852 | |

9 | 2 | 3 | 4 | 12 | 483 | 12×483=579618×297= 534627×198= 534628×157=4396 39×186=7254 42×138= 579648×159=7632 (Duplicate values are in bold and only one was kept for the sum of products.) | 45228 |

#### An initial solution

`print ([0,0,0,0,12,52,162,0,13458,45228][int(input())])`

#### A real–time solution

It's the same little brute–force hack I used to generate the list in the solution above. It runs in about 30ms for any chosen input and little effort in the way of optimization with one small exception. The product can never be more than four digits, so it must always be less than 9999. Except there are no valid factors that could create a product with a nine as the first digit so, I changed it to 8999. Not much, but passes all the HackerRank test cases.

#### The `is_pandigital()`

function

The `is_pandigital()`

function takes a string, `n`, and the intended length of that string, `s`, as arguments. The string represents a pandigital candidate with the designated length and is required to pass two tests to qualify as a 1 through `n` pandigital number:

- The length of the candidate string and the specified length,
`s`, are equal. A literal string, '1234567890', is truncated from the right by the number of characters specified in

`s`and all individual characters in`n`are stripped from the newly truncated string. Only when this resultant string is null is the candidate a pandigital number.For example, if n='12648' and s=5, then the literal '1234567890' is truncated to '12345'. Next, the characters from

`n`, '1', '2', '6', '4', '8', are stripped from '12345' leaving '35'. This would fail the pandigital test. However, a string like '24531' would pass.

#### HackerRank version

HackerRank Project Euler 32 extends the problem to include all 4–9 digit pandigital sets.

### Python Source Code

```
p = set()
def is_pandigital(n, s=9): return len(n)==s and not '1234567890'[:s].strip(n)
def calc (n):
for i in range(2, 99):
j = i+1
while i*j < 8999:
if is_pandigital(str(i)+str(j)+str(i*j), n): p.add(i*j)
j+= 1
return sum(p)
print (calc(int(input())))
```

### Last Word

- A Python
`set()`

was used to collect unique products and eliminate duplicates.