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View Full Version : Ok math gurus. See if you can figure out how this works.


Lzen
04-07-2005, 09:13 AM
http://digicc.com/fido/

Lzen
04-07-2005, 09:41 AM
C'mon CP, don't let me down. I need to know or I might go insane!


Well, more insane than I already am now. :p

Fezzic
04-07-2005, 10:02 AM
the 3 or 4 digits in your answer will string will always add up to a multiple of 9 (9, 18). So to figure out what digit is missing, just put in what would bring the sum up to a multiple of 9

for example, say the 4 numbers you get back in your answer are 3105 (as in one I tried). The sum of those 4 numbers is 9. if you tell it "105", that totals 6 and you know the missing number was a 3, etc.

if you give it 067, it will know the missing number is 6
if you give it 123, it will know the missing number is 3

(and if it doesn't work, your math is wrong) :)

-Fezzic

jspchief
04-07-2005, 10:06 AM
It got my number wrong

Lzen
04-07-2005, 10:52 AM
It got my number wrong

Yeah, I screwed it up when I started off with 999. :)

Lzen
04-07-2005, 11:01 AM
the 3 or 4 digits in your answer will string will always add up to a multiple of 9 (9, 18). So to figure out what digit is missing, just put in what would bring the sum up to a multiple of 9

for example, say the 4 numbers you get back in your answer are 3105 (as in one I tried). The sum of those 4 numbers is 9. if you tell it "105", that totals 6 and you know the missing number was a 3, etc.

if you give it 067, it will know the missing number is 6
if you give it 123, it will know the missing number is 3

(and if it doesn't work, your math is wrong) :)

-Fezzic

That's awesome. Thanks. BTW, I used 4743 and then 3744 with the result 999. Well, when you take one 9, you still have 99 left. Those add up to 18. So, what would you have guessed? Either a zero or 9? I assume this type of answer is the reason why it says don't circle a zero cause a zero is already a circle.

dtebbe
04-07-2005, 11:30 AM
Well this is a brute force type of proof restricted to 3 digit numbers. You'll see how to extend it to 4 digits but I won't do the whole thing.

For a 3 digit number (100x + 10y + z where x,y and z are digits from 0-9) there are 6 cases to consider:

case 1:
(100x+10y+z) - (100y +10z+x) = 99x-90y-9z = (9)(11x-10y-z) so it has a factor of 9. From there the rest follows from earlier e-mail.

case 2:
(100x+10y+z) - (100y +10x+z) = 90x-90y+0z = (9)(10)(x-y) again, factor of 9.

case 3:
(100x+10y+z) - (100z +10y+x) = 99x+0y-99z = (9)(11)(x-z) again, factor of 9.

case 4:
(100x+10y+z) - (100z +10x+y) = 90x+9y-99z = (9)(10x+y-11z) factor of 9.

case 5:
(100x+10y+z) - (100x +10z+y) = 0x+9y-9z = (9)(y-z) factor of 9.

case 6: this really isn't a case that is considered in the puzzle because you are required to mix up the numbers.
(100x+10y+z) - (100x +10y+z)=0x+0y+0z = 0.

In general, then number of cases is (number of digits) "factorial" - 1 (the zero case) OR 3! - 1 = 6 - 1 = 5.
In all allowed cases, the difference has a factor of 9, therefore is divisible by nine, therefore the sum of the digits is divisible by nine etc.
Now no restriction is made here for positive or negative numbers. The difference could be negative in general, and you could still work things out. You would just need to know more information (namely was the difference positive or negative) to figure out what the person circled. The point is, if the proof is true in general, it is certainly true if you restrict it to differences between two numbers where the first is larger. I'm not sure that you care, but here is a case where it is easier to prove the general case (+ and -) and show that the specific case (+) is a subset than to prove just that (+) is true.

For a four digit number the proof would be the same except that the number of cases to consider is 4!-1=24-1=23 cases.


DT