38 lines
1.2 KiB
Ruby
38 lines
1.2 KiB
Ruby
descr=%{
|
||
Take the number 192 and multiply it by each of 1, 2, and 3:
|
||
|
||
192 × 1 = 192
|
||
192 × 2 = 384
|
||
192 × 3 = 576
|
||
|
||
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
|
||
|
||
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
|
||
|
||
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
|
||
}
|
||
|
||
$numbers=%q(123456789).split('').sort
|
||
def is_pandigital(str)
|
||
return str.split('').sort == $numbers
|
||
end
|
||
|
||
best=(123456789)
|
||
(2..10).each do |n|
|
||
concat_prod=""
|
||
base=1
|
||
while concat_prod.length < 10
|
||
concat_prod=""
|
||
(1..n).each do |i|
|
||
concat_prod <<= (base*i).to_s
|
||
end
|
||
if is_pandigital(concat_prod)
|
||
puts %{base=#{base} n=#{n} #{concat_prod}}
|
||
if concat_prod.to_i > best
|
||
best=concat_prod.to_i
|
||
puts %{* base=#{base} n=#{n} #{best}}
|
||
end
|
||
end
|
||
base += 1
|
||
end
|
||
end
|