[pineapplejellynotloggedin]

Gah!!! I solved this ages ago and have been trying at regular intervals over the past six MONTHS to a YEAR to find a second solution! I wish I'd known I didn't need it!!!

[X9Tim]

This card was in my first pack, and I tried to finish them all before going online.

I took it as written, i.e. "There are two possible solutions to this, what are they?"

I got one solution fairly quickly, but I was getting tired and kept making the same silly mistake.... :s

After a troubled night, I got the second answer next morning and went online to enter it. I was infuriated to find I only needed one answer!

I think this counts as a printing error!!

--
Still I've got the points, and that's what matters!

[extrusion]

I just used Excel with a COUNTIF function, after sorting out the circular references stuff it complains about, the answers just appear like magic!

extrusion

[jayito]

I found this one really obvious and really easy, which was a shock as I normally find the cards quite hard! It's nice to know I am not completely dense!

[kyz]

Just in case there were more solutions, I wrote a program to exhaustively search the problem space. It checks every possibility from 1 1 1 1 1 1 1 1 1 1 to 14 14 14 14 14 14 14 14 14 14.

Code:

#include <stdio.h> #define FOR(x) for (x = 1; x <= 14; x++) #define ADD(x) if (x > 9) { n[x/10]++; n[x%10]++; } else { n[x]++; } int main() {   int c[10], n[10];   FOR(c[9]) FOR(c[8]) FOR(c[7]) FOR(c[6]) FOR(c[5])   FOR(c[4]) FOR(c[3]) FOR(c[2]) FOR(c[1]) FOR(c[0])   {     n[0] = n[1] = n[2] = n[3] = n[4] = n[5] = n[6] = n[7] = n[8] = n[9] = 1;     ADD(c[0]); ADD(c[1]); ADD(c[2]); ADD(c[3]); ADD(c[4]);     ADD(c[5]); ADD(c[6]); ADD(c[7]); ADD(c[8]); ADD(c[9]);     if (c[0]==n[0] && c[1]==n[1] && c[2]==n[2] && c[3]==n[3] && c[4]==n[4] &&         c[5]==n[5] && c[6]==n[6] && c[7]==n[7] && c[8]==n[8] && c[9]==n[9])       printf("SOLUTION: %d %d %d %d %d %d %d %d %d %d\n",         c[0],c[1],c[2],c[3],c[4],c[5],c[6],c[7],c[8],c[9]);   }   return 0; }

Even with this search space, there are still only 2 solutions:

Spoiler (Rollover to View):

SOLUTION: 1 11 2 1 1 1 1 1 1 1 SOLUTION: 1 7 3 2 1 1 1 2 1 1

If someone wants to try a larger search space (e.g. 1 to 22), there might be something... but I doubt it.

[sabsay]

Spoiler (Rollover to View):

1,7,3,2,1,1,1,2,1,1 is the correct answer

[perplextown]

NEED HELP!!!!

[Hunting4Treasure]

ReeKorl wrote:

Not sure how relevant this is: Spoiler (Rollover to View): but has anyone else noticed the single digit answer (1732111211) is the same as the first ten digits of the square root of three?

I thought that was way cool, and was getting ready to try different starting points to see what I'd come up with. But, what I got for an answer is:

Spoiler (Rollover to View):

1.7320508075688772935274463415059 making the first 4 numbers the same, but then it falls apart from there. 1732111211 squared, actually turns out to be 3.000209247271886521. Now, the first *5* numbers equal 3 (which is cool enough), but not all 10 as stated above.

Am I missing something? Or, is *that* sentence *also* false?

[Schmooze]

Third Solution

[looosy]

Quote:

has anyone else noticed the single digit answer (1732111211) is the same as the first ten digits of the square root of three?

Interesting! so many things keep pointing back to 3 and cubed....

Very clever

[ReeKorl]

Not sure how relevant this is:

Spoiler (Rollover to View):

but has anyone else noticed the single digit answer (1732111211) is the same as the first ten digits of the square root of three?

[JebJoya]

interestingly, the site only asks for one solution to give you the points, even though the card clearly asks for both... hey ho...

Jeb

[RobDixon]

lauriek,

Thanks for posting the solver programs!

I really like the ONE answer, but you're right, I think it won't count since the rules say to add digits.

Here's another variation that only involves digits. It seems to fit the rules, though it's not quite as simple as the first two solutions:

Spoiler (Rollover to View):

THE NUMBER OF TIMES THE DIGIT 0 APPEARS IN THIS PUZZLE IS 10 THE NUMBER OF TIMES THE DIGIT 1 APPEARS IN THIS PUZZLE IS 11 THE NUMBER OF TIMES THE DIGIT 2 APPEARS IN THIS PUZZLE IS 02 THE NUMBER OF TIMES THE DIGIT 3 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 4 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 5 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 6 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 7 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 8 APPEARS IN THIS PUZZLE IS 01 THE NUMBER OF TIMES THE DIGIT 9 APPEARS IN THIS PUZZLE IS 01

Just doing my bit to help sow the seeds of confusion.

[lauriek]

Oh and another /possible/ answer

Spoiler (Rollover to View):

the number of times the digit 0 appears on this puzzle is ONE the number of times the digit 1 appears on this puzzle is ONE the number of times the digit 2 appears on this puzzle is ONE the number of times the digit 3 appears on this puzzle is ONE the number of times the digit 4 appears on this puzzle is ONE the number of times the digit 5 appears on this puzzle is ONE the number of times the digit 6 appears on this puzzle is ONE the number of times the digit 7 appears on this puzzle is ONE the number of times the digit 8 appears on this puzzle is ONE the number of times the digit 9 appears on this puzzle is ONE But I think this is excluded by the use of the word DIGIT in the question..

[lauriek]

I don't think a blue card is going to be that easy, the others I have certainly don't seem to be!

Spoiler (Rollover to View):

I wrote these two scripts which come up with the same answers as helper... basic stepper script randomized stepper Both pages use some pretty simple javascript to do the stepping... (note its NOT the randomizer button on the second script which makes it find the second answer, its a random element in the 'update number when its wrong' routine which only updates the number 50% of the time it's wrong)