Arrange the digits from 0 to 9 into a number which is divisible by every integer from 12 to 21 inclusive.
(There are 2 right answers.)
$\begingroup$
$\endgroup$
asked Aug 16, 2016 at 3:35
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
The numbers are
2137458960 and 6179584320
I found these results by calculating the least common multiple $\operatorname{lcm}(12,\dotsc 21) = 21162960$, then checking all multiples of this value up to 9876543210 with a script: https://jsfiddle.net/qdnyLfx8/
There must be more elegant solutions that work without using a computer.
-
1$\begingroup$ Well then I won't post the 2nd number. $\endgroup$ Commented Aug 16, 2016 at 4:11
-
$\begingroup$ The prime factors of 12 to 21 are 24×3×5×7×13×17×19, therefore the two 10-digit pandigital numbers are a multiple of 7054320. 2137458960 [7054320×303] and 6179584320 [7054320×876]. $\endgroup$ Commented Aug 16, 2016 at 4:29
-
1$\begingroup$ @JamalSenjaya Based on GOTO 0's LCM calculation you seem to have missed a factor of 3 somewhere. $\endgroup$ Commented Aug 16, 2016 at 8:46
$\begingroup$
$\endgroup$
If I take all the factors of 12-21 and multiply them together, I get
5 * 7 * 9 * 13 * 16 * 17 * 19, or
21162960
So now all that is left to do is to keep adding it to itself until I get a 10 digit number that satisfies the requirements. The first such number is
2137458960
answered Aug 16, 2016 at 4:02