Pandigital multiples

2014-09-24

Problem 038: Pandigital multiples

Description:

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?

Solution:

v#########
v#########    v                                                                                                                                              <#     p145<
#########         v          +1<           v            <  v               <                 vp2\0:<         >v    >v          >$v   v\g2:<
>"ec"*31p>241p>01-1>:31g*\:41g-#^_$>\10p01-\>:55+%\55+/:#^_$>10g55+*+10p:1+#^_$10g\:1+#v_$:55+>  1-:|  >:55+%:|>:2g!|>1\2p55+/:!| >55+> 1-:|>#v_$41g1-:41p1-#^_31g1-:31p|
                                   ^                                                  \<           $>$:^      $#    $#          <v++++++++$<
                                                                                                            $$>     >          $0>9-!\$     ^ >.     @     ,,,,, "RORRE"<

Start

Pause

Reset

Download

Output:

Stack: (0)

Explanation:

Not much to say here. I needed to implement a way of merging numbers and a way of testing for pandigitals (we already did that in problem 32). Then we start from 9999 downward, because bigger numbers will always lead to 10 digits or more.

Interpreter steps:	3 567 967
Execution time (BefunExec):	624ms (5.72 MHz)
Program size:	169 x 6
Solution:	932718654
Solved at:	2014-09-24