Slightly shorter if the formatted results are not converted back into
numbers:
>./ (#~ (-: |.)@("."0@":)"0) , */~ 100+i. 900
906609
>./ (#~ (-: |.)@":"0) , */~ 100+i. 900
906609
A different and perhaps less brutish method is to start with 6-digit
palindromes and see which ones have two 3-digits factors.
On Sun, Jun 8, 2014 at 8:00 AM, Henry Rich <[email protected]> wrote:
> Brute force seems right for this problem. You could write your solution
> as
>
> >./ (#~ (-: |.)@("."0@":)"0) , */~ 100+i. 900
>
> Henry Rich
>
> On 6/8/2014 10:46 AM, Jon Hough wrote:
>
>> I am pretty pleased that I completed Project Euler Q4.
>> Question:
>>
>>
>> A palindromic number reads the same both ways. The largest palindrome
>> made from the product of two 2-digit numbers is 9009 = 91 99.
>>
>> Find the largest palindrome made from the product of two 3-digit numbers.
>>
>>
>> https://projecteuler.net/problem=4
>>
>>
>> My solution:
>>
>> base =. "."0 ":
>>
>> NB. Tests if y value is palindrome.
>> is_palindrome =. (# =)@:(=|.)@:base
>> NB. multiply all 3 digit nums (100 ~ 999)
>> mult =: (* is_palindrome"0) @ (*/)
>> NB. Create list of 3 digit nums.
>> list =. 100+ i.900
>>
>> (>./)@:, list mult list
>> Although I'm glad got the answer, I'm wondering if it could be made
>> terser, or if there is a much terser way to solve it? I went for a brute
>> force approach.
>> For comparison here is a Haskell answer I found:
>> maximum[a*b|a<-[100..999],b<-[a..999],reverse(show(a*b))==show(a*b)]
>> Regards.
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