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NOTE: Here confusion refers to the fact that the number if written on a piece of paper would appear as different number if viewed upside down. 18 may be viewed as 81 and so on

Can anyone find a generalized way to calculate the number of numbers which cause/don't cause confusion when rotated?

For $2$ digits... Total possible numbers... $9\times 10=90$ Digits causing confusion $= 4 (1,6,8,9)$ Possible numbers.... $4\times 4 = 16$ (Number can be confusing IFF both the digits are confusing, $0$ not considered because the number can't be single digit) Exceptions... $11, 69, 88, 96$

Soln... $90-16+4= 78$

Is there a generalized solution for $3$ digits? Because the exceptions would be much harder to find.. Number of possible confusions for 3 digits are... $4\times 5\times 4 = 80$ ($0$ can't be first or last)

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  • $\begingroup$ For abc to be an exception, you need $a=c=1$ or $a=c=8$ or $a=6,c=9$, or $a=9,c=6$, and you need $b=0$ or $b=1$ or $b=8$, right? $\endgroup$ Commented Jun 25, 2014 at 11:37
  • $\begingroup$ Yes. That makes 12 such combinations. Basically what I've considered is that the symmetry would be calculated as the number of digits. For 4 digits, a==d and b==c combinations, which should put the exceptions at 16. Is this line of thinking correct? $\endgroup$
    – Arcane
    Commented Jun 26, 2014 at 3:38
  • $\begingroup$ For 4 digits, ad could be 11, 88, 69 or 96; bc could be any one of those, or 00, so I get 20 exceptions. $\endgroup$ Commented Jun 27, 2014 at 6:44
  • $\begingroup$ So are we on the same page? $\endgroup$ Commented Jun 29, 2014 at 5:58
  • $\begingroup$ Seems like it.. More of a smmetry question than permuting them. But thanks all the same $\endgroup$
    – Arcane
    Commented Jun 30, 2014 at 3:21

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