
Anyone Know How They Do This?

RE: Anyone Know How They Do This?
Ok, I've got this far:<BR>"Choose any 4digit number":<BR>If we choose "abcd", that's actually:<BR> 1000a + 100b + 10c + d<BR>Jumble those up and you get another 4digit number:<BR> 1000e + 100f + 10g + h<BR>Subtract them:<BR>=> 1000(ae) + 100(bf) + 10(cg) + (dh)<BR><BR>Now, I'm having trouble. I think I can then get rid of a lot of those values because of the fact that efgh is a jumble of abcd...<BR><BR>Craig.

ahh the wonderful world of math
i don't have time to figure it out or else i would be. there's more math behind it than that stupid one, if you've seen it you'll know, but it's like a crystal ball with a bunch of symbols on the right. you pick a number, do a calculation, and find your symbol, then rub the crystal ball and voila! the symbol that you picked out is shown in the crystal ball. it's stupid. anyway... i'll post the algorithm :P probably not but here's to optimism.

RE: ahh the wonderful world of math
yeah, i figured it was some kind of mathematical algorithm, etc., but i just wanted to see if anyone here knew of a link to it.<BR><BR>thanks.

RE: Anyone Know How They Do This?

Oh, come on! It's just math!
Try this one:<BR><BR>Pick a 3 digit number; don't make 1st and 3rd digits the same.<BR><BR>Reverse that number to get a second number.<BR><BR>Subtract the smaller from the larger to get a 3rd number. (If less than 3 digits, zerofill on the left.)<BR><BR>Reverse *that* result to get a 4th number.<BR><BR>Add the 3rd and 4th numbers.<BR><BR>Your answer is 1089.<BR><BR>Same principal as what they are doing.<BR><BR>Mine is easy to see mathematically:<BR><BR>ABC is really A*100 + B*10 + C<BR>reverse that and the number is C*100 + B*10 + A<BR><BR>Subtract those:<BR> (A*100 + B*10 + C)  (C*100 + B*10 + A)<BR>and you get<BR> A*99  C*99<BR>or<BR> 99 * ( A  C )<BR><BR>So the only possible answers from the subtraction are 099, 198, 297, 396, etc.<BR><BR>And those results are all of the form<BR> X*100 + 90 + (9X)<BR>reverse that and you get<BR> (9X)*100 + 90 + X<BR>add those and you get<BR> (9X+X)*100 + 180 + (9X+X)<BR>which is of course just<BR> 900 + 180 + 9<BR>which is 1089.<BR><BR>So they are just doing something similar to this. Probably based on "casting out nines."<BR><BR>There's no "trick" involved. Just have to spend the time to analyze the simple arithmetic involved.<BR><BR>

Yeah, it's easy to show...
...that after the first step, the result of the subtraction will give you an answer where the digits all add up to a multiple of 9:<BR><BR>Example: 715 first number<BR>715  157 = 558 (sum of 5+5+8 is 18)<BR>715  175 = 540 (sum 9)<BR>715  517 = 198 (sum 18)<BR>715  571 = 144 (sum 9)<BR>751  715 = 36 (sum 9)<BR><BR>So when you circle (omit a number) from any of those sums and tell him the other digits, he just has to add some number to make the total be a multiple of 9.<BR><BR>Presto. Dirt simple.<BR><BR>Proving that all possible scramblings will result in a number where the sum of the digits is 9 will take a little bit of thought, but it's not rocket science. Hell, for any 3 digits, worst comes to worst you just construct all of the 5 possible subtractions to show it. Or for 4 digits, all of the 23 possible subtractions. But I'm sure there's an easier way.<BR><BR>I think my 1089 trick is more impressive than that one.<BR><BR>

Yep, pretty simple..
just one of those multiple of 9 things..<BR><BR>mostly the answer will add up to 18<BR>eg 4086, 5265, 3474 etc...<BR>so its a simple matter if division from 18 of the numbers you enter to find out what you circled.<BR><BR>Of course your answer could add up to 9 <BR>e.g. 0900 <BR>not too difficult to get.

OK, 1 min + a few secs behind
but I only just got in to work, still fumbling for that first cup.

I think you are on right track...
Worst come to worst, try all possible combinations!<BR><BR>To pick just one example:<BR><BR>Scramble abcd to dbca:<BR> 1000(ad) + 100(bb) + 10(cc) + (da)<BR>which is of course<BR> 1000(ad)  (ad)<BR>result<BR> 999(ad)<BR>which is patently a number where the digits add up to 9!<BR><BR>Scramble abcd to bacd:<BR> 1000(ab) + 100(ba) + 10(cc) + (dd)<BR>result<BR> 900(ad)<BR>ditto on the digits!<BR><BR>Gets more complex when you scramble 3:<BR>Scramble abcd to bcad:<BR> 1000(ab) + 100(bc) + 10(ca) + (dd)<BR>result<BR> 1000a10a + 100b1000b + 10c100c<BR>which is<BR> 990a  900b  90c<BR>ditto on the digits!<BR><BR>So it's not really complex at all!<BR><BR>*ALL* of the subtractions end up with the multiplier of each digit being itself a multiple of 9!<BR><BR>
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