Attached Link: http://digicc.com/fido/

doesn;t work for me with 9546 mixed up: 4659

it seems that the difference between a number and its anagram is always a multiple of 9. (for example 91 and 19 are anagrams, their difference is 72, which is a multiple of 9 ... try it for other numbers). now, i seem to remember that a number is divisible by 9 only if the sum of its digits divisible by 9. Take 36, obviously divisible by 9. The sum of its digits 3 and 6 is 9, which is divisible by 9. Or try 99: sum of the digits is 18, also divisible by 9. So when you type in your three non-circled digits (or two if you started with a three digit number), it adds them up and then figures out what you'd have to add to that sum to get a multiple of 9. this is the number it guesses. for example for my number 9441, the anagram was 4914 and the difference 4527 (which is 9*503). if i circle 7, i'd type in 452, it adds the digits (4+5+2) to get 11. since it needs 7 more to make a number divisible by 9 (namely 11+7=18) it correctly guesses 7.

A number minus its anagram is divisible by nine, like so: Take the number XYZT where the letters represent numbers. This is the same as X1000 + Y100 + Z10 + T However you rearrange the numbers (anagram), you will end up with the individual numbers times a coefficient divisible by nine or zero. Because it will either be zero (when the position of the number hasn't moved) or the difference between two powers of ten (10^a - 10^b). Example, if we rearrange the number above to read ZTYX, the difference will be: X(1000-1)+Y*(100-10)+Z*(10-1000)+T*(1-100)= X999 + Y90 + Z*(-990) + T*(99) where all of the coefficients are divisible by nine. Which makes the products with X,Y etc. divisible by nine, which makes the sum of those products divisible by nine.. The negatives do not matter, since even if they are negative, they are STILL divisible by nine..

Similarly, a number can only be divisible by nine if the digits add up to nine, like so: If we take the number from the previous proof to use as an example, XYZT is X1000 + Y100 + Z10 + T Let us examine the remainders if we divide all of these individual components by nine... X1000 / 9 = (actually X1000 mod 9) = X * (999+1) / 9 = 111X with the remainder of X similarly for Y,Z and T. So we know XYZT is divisible by nine, if the sum of the remainders are divisible by nine as well. Well, since the remainders are the original digits themselves, that proves that any number of any length is divisible by nine, if the digits sum up to nine.

"http://digicc.com/fido/ formula" If you can type it, you can google it.

Wow thanks for pointing it out! I've done work with the X(1000-1)+Y*(100-10)+Z*(10-1000)+T*(1-100)= X999 + Y90 + Z*(-990) + T*(99)* formula, so I've got it all figured out. Pretty inventive for an advertisement, I'd say. Maybe I'll forward it to the Math teacher, mooching extra credit is not an exact science. Thanks!

You guys are making my head hurt. I barely have the hang of 2 + 2 = 4.

yeah, senic and moongoat are wise!

woah awesome stuff. And xenic dude, thanx for the explaination! :D

nah, this is the other thing I learned in 2 semesters of Accounting..if you transpose a number, the difference is divisible by 9. (the other is using your knuckles to find months with 31 day..silly trick, but it works..;)

hey, that knuckle trick I did know! lol a teacher told us that in primary school here.