7
$\begingroup$
 _ _ _ _ _ × 8 = _ _ _ _ _

There is a number that you can put in the left-hand side of the above equation so that the blanks are filled in using every digit from 0 to 9 exactly once. What is that number?

$\endgroup$
3
  • 3
    $\begingroup$ If there are multiple solutions, you are usually expected to state that. $\endgroup$ Nov 5 '14 at 16:41
  • $\begingroup$ I didn't actually know there were multiple solutions myself. I was a bit hasty posting this puzzle, it seems. $\endgroup$
    – user88
    Nov 5 '14 at 16:43
  • $\begingroup$ how to find the number of solutions of such a pandigital puzzle? $\endgroup$
    – manshu
    Mar 6 '16 at 13:22
18
$\begingroup$

There is more than one correct answer.

(10459, 83672), (10469, 83752), (10537, 84296), (10579, 84632), (10592, 84736), (10674, 85392), (10679, 85432), (10742, 85936), (10794, 86352), (10932, 87456), (10942, 87536), (10953, 87624), (10954, 87632), (12073, 96584), (12307, 98456), (12345, 98760)

$\endgroup$
3
  • $\begingroup$ Huh, I had just gotten to your third answer right before this showed up. $\endgroup$
    – JonTheMon
    Nov 5 '14 at 15:26
  • $\begingroup$ +1. Thanks, I tried to find a proof for the JonTheMon's solution and failed to do so. What an ugly tricky puzzle... $\endgroup$
    – klm123
    Nov 5 '14 at 16:38
  • $\begingroup$ I was afraid this would be the case. $\endgroup$
    – user88
    Nov 5 '14 at 16:38
6
$\begingroup$

Oddly enough, the numbers are 12345 & 98760. I knew that the first two digits of the top had to be low numbers, 1 and 2, so I just tried it out, and it worked!

This worked out well since the 1's digit of the multiplication (before carrying) ends up being 86420 (8 stepping down by 2), and we have carry values of 12340
So, we get:
8 - 0 + 1 = 9
8 - 2 + 2 = 8
8 - 4 + 3 = 7
8 - 6 + 4 = 6
8 - 8 + 0 = 0

$\endgroup$
2
  • $\begingroup$ Yep, that's correct! $\endgroup$
    – user88
    Nov 5 '14 at 14:53
  • 1
    $\begingroup$ 0 is lower than 2... the complexity of this puzzle is to find the proof, not to find the answer.. $\endgroup$
    – klm123
    Nov 5 '14 at 15:11
4
$\begingroup$

A list of all the possible solutions to this:

(03187, 25496)
(04589, 36712)
(04591, 36728)
(04689, 37512)
(04691, 37528)
(04769, 38152)
(05237, 41896)
(05371, 42968)
(05789, 46312)
(05791, 46328)
(05839, 46712)
(05892, 47136)
(05916, 47328)
(05921, 47368)
(06479, 51832)
(06741, 53928)
(06789, 54312)
(06791, 54328)
(06839, 54712)
(07123, 56984)
(07312, 58496)
(07364, 58912)
(07416, 59328)
(07421, 59368)
(07894, 63152)
(07941, 63528)
(08174, 65392)
(08179, 65432)
(08394, 67152)
(08419, 67352)
(08439, 67512)
(08932, 71456)
(08942, 71536)
(08953, 71624)
(08954, 71632)
(09156, 73248)
(09158, 73264)
(09182, 73456)
(09316, 74528)
(09321, 74568)
(09352, 74816)
(09416, 75328)
(09421, 75368)
(09523, 76184)
(09531, 76248)
(09541, 76328)
(10459, 83672)
(10469, 83752)
(10537, 84296)
(10579, 84632)
(10592, 84736)
(10674, 85392)
(10679, 85432)
(10742, 85936)
(10794, 86352)
(10932, 87456)
(10942, 87536)
(10953, 87624)
(10954, 87632)
(12073, 96584)
(12307, 98456)
(12345, 98760)

Don't forget you can use 0 at the start of the first number :)

$\endgroup$
1
  • $\begingroup$ usuallly in puzzles of this type leading zero is disallowed. $\endgroup$
    – Jasen
    Jun 18 '16 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy