Wed 13 Aug, 2003 06:14 am
Not sure why it works, Kev -- but if you give me the numbers you would normally write in the blank -- I will tell you what number you circled.
I'm on to you frank, the first thing that I see is that when you add the two numbers together and subtract them from the original number they are always exactly divisibe by 9. I haven't worked my riddle out yet, but I'll get there eventually.
The amount of guiness I imbibe doesn't seem to be much help.
The number 9 is significant in the solution to your puzzle also -- but not in the same way.
When you choose a 4 digit number, numerals in the answer will ALWAYS add up to 9, 18, or 27.
EXAMPLES: 8147-1478=6669 (6+6+6+9=27)
Now as you can see, ALL of the answers have a 6
in them. So how does it decipher that you circled a 6 since the 4 digits don't add up to the same number everytime? Okay...look at the first one. If you circled a "6", the remaining numbers add up to what? "21"...correct? That's more than "18" and "9" RIGHT THERE. So it knows that in THIS particular number, the numbers added up to 27. So knowing THAT, and knowing that the numbers you typed in add up to "21", WHAT NUMBER plus "21" equals 27?
SIX!! What number did you circle? SIX!! Now look at the one where the answer added up to "9".
(2014-1402=0612). Circle the "6" in THIS one and what numbers are left? "012". What do those numbers add up to?....THREE. So it knows that in THIS particular number, the 4-digits added up to "9". How? What number plus 3 equals 18? FIFTEEN.....a TWO DIGIT NUMBER. You can't circle 2 numbers. Same ordeal with the number 27. So knowing that this number is gonna add up to "9", and knowing that the numbers you typed in add up to "3", what number plus 3 equals 9?....SIX!! What number did you circle?....SIX!! That's how this mind bender works.
Welcome to Able2Know sbrean11!
As long as you follow the one essential rule (ie, do not circle a zero) then it doesn't matter what digit or digits you enter into the box, nor in what order. The program will divide your entry by 9 and any resulting remainder will be subtracted from 9 to determine the number you chose. If the remainder is 3, then the number you dropped from your entry is 9 - 3 = 6. A remainder of 5 would indicate you selected 4. If there is no remainer, because 9 divides evenly into your entered number, then you chose 9 - 0 = 9.
If you are naughty and choose a zero, the program will be foiled. For instance, if the difference between your original number and the number as jumbled amounts to 4067 and you enter 647 instead of, say 670, the program will assume you chose a 1 rather than zero. 647 / 9 = 71 with a remainder of 8, and 9 - 8 = U-no, what?
When you jumble your original number, you are in essence transposing digits. And any old bookkeeper who's ever lost his/her balance knows to divide the amount of the imbalance by 9 (casting out nines, they call it) to see if perchance he/she has transposed digits somewhere, as opposed to having made an arithmetical boo-boo.
The puzzle here is like dropping a digit from the "bookkeeper's" imbalance, making it not evenly divisible by 9. And as long as the "circled" digit is not a zero, there will be a remainder, which subtracted from 9, will reveal the missing digit. Alternatively, you can mentally place on either side of the remainder the digit which will make up a multiple of 9. That is, if the remainder is 2, then placing a 7 before or after it, you will have a multiple of 9 ( 72 or 27 ) and 7 would therefore be the chosen digit. Simpler though to just subtract the remainder from 9.
I was a bit hesitant about posting this because I figgered I'd end up making a meal of it. So I'll shut up after mentioning that it doesn't matter how many digits the original number has, as long as it has at least two. Kinda hard to transpose a number that has only one digit. Also, it's not necessary to do the second shuffle after selecting the digit not to be typed into the box. That's just padding.