All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Pandigital and divisible (Posted on 2010-12-26)
What percentage of pandigital* numbers is divisible by 990 ?

*A pandigital number consists of 10 distinct digits.

Try to solve analytically.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solution | Comment 1 of 4

I couldn't think of any analytic ways: All are divisible by 9, but only some are divisible by 2, 5 or 11. It's especially tricky if ruling out leading zeros. So falling back on the computer:

`   10   V\$="0123456789":H\$=V\$   15   repeat   20    gosub *Permute(&V\$)   30    if left(V\$,1)>"0" then   40      :inc OCt   50      :if val(V\$) @ 990=0 then inc HCt  735   until V\$=H\$  740   print HCt,OCt,HCt/OCt,HCt//OCt  750   end  `

finds those cases where there is no leading zero. The results are:

31680   3265920         0.0097001763668430334   11//1134

showing the probability as 31680/3265920 11/1134 ~= 0.97 %, or 1/103.090909090909090909.

The 3265920 is 10! - 9!, representing all cases, including leading zeroes, minus the cases with leading zeroes.

However, if leading zeros are allowed, then the following program takes into consideration all 10! = 3,628,800 cases:

`   10   V\$="0123456789":H\$=V\$   15   repeat   20    gosub *Permute(&V\$)   30    if left(V\$,1)>" " then   40      :inc OCt   50      :if val(V\$) @ 990=0 then inc HCt  735   until V\$=H\$  740   print HCt,OCt,HCt/OCt,HCt//OCt  750   end  finds`

31680   3628800         0.0087301587301587301   11//1260

That's 11/1260 ~= 1/114.5454545454545454545.

 Posted by Charlie on 2010-12-26 19:45:19

 Search: Search body:
Forums (0)