Any Math specialists out there???

1fog · #1 03-23-2006, 11:11 PM

http://digicc.com/fido/

Brought to you by 7up. I just cant figure out how they do it. Any info would be appreciated. Click on the kid in the bottom right get it started.

pzhon · #2 03-23-2006, 11:40 PM

When you scramble the digits of a number, you don't change the remainder when you divide by 9. Subtracting the scrambled number from the original produces a number divisible by 9, whose digits must add up to a multiple of 9. If you type in digits that add up to 4 more than a multiple of 9, the missing digit must be 5.

There is some ambiguity. If you type in digits that add up to a multiple of 9, the missing digit can be 0 or 9. I believe the site will guess 9, as this is more likely if you have followed the other instructions.

1fog · #3 03-23-2006, 11:46 PM

[ QUOTE ]
When you scramble the digits of a number, you don't change the remainder when you divide by 9. Subtracting the scrambled number from the original produces a number divisible by 9, whose digits must add up to a multiple of 9. If you type in digits that add up to 4 more than a multiple of 9, the missing digit must be 5.

There is some ambiguity. If you type in digits that add up to a multiple of 9, the missing digit can be 0 or 9. I believe the site will guess 9, as this is more likely if you have followed the other instructions.

[/ QUOTE ]

wow, xcllnt!!!

shermanash · #4 03-23-2006, 11:48 PM

[ QUOTE ]

There is some ambiguity. If you type in digits that add up to a multiple of 9, the missing digit can be 0 or 9. I believe the site will guess 9, as this is more likely if you have followed the other instructions.

[/ QUOTE ]

they get around this by saying you can't circle (eliminate) 0 because it "already is a circle"

drbst · #5 03-23-2006, 11:49 PM

EDIT: Oops, too late. [img]/images/graemlins/smile.gif[/img]

If you rearrange the digits of a number to construct another number and subtract both from each other, the checksum is always a multiple of 9.

Therefore, the missing number must be the difference between the checksum and the integer that is a multiple of 9 and that is larger than the checksum.