_ _ _ _ _ × 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?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ If there are multiple solutions, you are usually expected to state that. $\endgroup$– Ross MillikanNov 5, 2014 at 16:41
-
$\begingroup$ I didn't actually know there were multiple solutions myself. I was a bit hasty posting this puzzle, it seems. $\endgroup$– user88Nov 5, 2014 at 16:43
-
$\begingroup$ how to find the number of solutions of such a pandigital puzzle? $\endgroup$– manshuMar 6, 2016 at 13:22
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
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)
-
$\begingroup$ Huh, I had just gotten to your third answer right before this showed up. $\endgroup$ Nov 5, 2014 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... $\endgroup$– klm123Nov 5, 2014 at 16:38
-
$\begingroup$ I was afraid this would be the case. $\endgroup$– user88Nov 5, 2014 at 16:38
$\begingroup$
$\endgroup$
2
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
-
$\begingroup$ Yep, that's correct! $\endgroup$– user88Nov 5, 2014 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.. $\endgroup$– klm123Nov 5, 2014 at 15:11
$\begingroup$
$\endgroup$
1
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 :)
-
$\begingroup$ usuallly in puzzles of this type leading zero is disallowed. $\endgroup$– JasenJun 18, 2016 at 1:10