Puzzle: Numbers on BlackBoard

[Eccentric]

The teacher put up some digit on blackboard.
To be precise teacher puts exactly twenty '1's, Thirty '2's and Thirty five '3's on the board.

then she asks randomly picks one student and him to perform exctly one out of following three operations (if possible):
1. Erase one '1' and one '2' and put an extra '3'.
2. Erase one '2' and one '3' and put an extra '1'.
3. Erase one '3' and one '1' and put an extra '2'.

She keeps on picking students and doing this and number of digits on the board keeps on reducing.
What can be said about the last digit on the board? If it will be a particular one which will be the last digit remaing on the board? else what all digits can be the last remaining on the board and why???

[nzomniac]

It will always be 3
Look below for Matlab code
clear all;
x=20;
y=30;
z=35;
while(x+y+z>1)
choice=rand(1);
if (choice<0.3333)&(x~=0)&(y~=0)
x=x-1;
y=y-1;
z=z+1;
end;
if (choice>0.3330)&&(choice<.6666)&(z~=0)&(y~=0)
y=y-1;
z=z-1;
x=x+1;
end
if (choice>.6666)&(z~=0)&(x~=0)
x=x-1;
z=z-1;
y=y+1;
end

end

This always terminates with x=0;y=0;z>1

[jockey]

it's 10 2's and 15 3's,

after one full operation is done i.e all three steps, no. of 1 will be reduced by 1 as i have erased it twice and written it once, same will happen with 2 and 3

so at the end of 20 operations, no. of 1,2 and 3 will be reduced by 20. and i will have only 10 2's and 15 3's left, now i can't do step no. 1 as i don't have any 1 to erase.

so finally left no. would be ten 2's and fifteen 3's


am i right eccentric

[Eccentric]

com'on you don't need a mat-lab code to settle that..
come out with some ogic dude!!

[Eccentric]

Originally Posted by jockey

it's 10 2's and 15 3's,

after one full operation is done i.e all three steps, no. of 1 will be reduced by 1 as i have erased it twice and written it once, same will happen with 2 and 3

so at the end of 20 operations, no. of 1,2 and 3 will be reduced by 20. and i will have only 10 2's and 15 3's left, now i can't do step no. 1 as i don't have any 1 to erase.

so finally left no. would be ten 2's and fifteen 3's


am i right eccentric

one operation doesn't mean carry out all three steps.... pick any one out of three...
SO even at the stage you specified we can proceed by removing one 2 and one 3 and putting a 1 instead !

[DesiGuru]

3 is the right answer!
whatever you do will be left with a 1 and a 2 in the end which will lead to 3 as the end result. You can try doing it manually by taking the best and worst cases...
The theory and logic is beyond me right now... I'll see if i can figure it out...

K

[khanna_sumit]

Originally Posted by Eccentric

The teacher put up some digit on blackboard.
To be precise teacher puts exactly twenty '1's, Thirty '2's and Thirty five '3's on the board.

then she asks randomly picks one student and him to perform exctly one out of following three operations (if possible):
1. Erase one '1' and one '2' and put an extra '3'.
2. Erase one '2' and one '3' and put an extra '1'.
3. Erase one '3' and one '1' and put an extra '2'.

She keeps on picking students and doing this and number of digits on the board keeps on reducing.
What can be said about the last digit on the board? If it will be a particular one which will be the last digit remaing on the board? else what all digits can be the last remaining on the board and why???

I believe that the answer lies in the no of these no.s being even or odd...lets call it polarity...

initially no. of 1s and 2s are even and no of 3s are odd......
lets analyse the three operations
1. -(1), -(2), +(3) so if polarity is even even odd initially then polarity becomes odd odd even...or lets say eeo---ooe
2. +(1), -(2), -(3) means eeo----ooe
3. -(1), +(2), -(3) means eeo----ooe

similarly in any case if we start with ooe, we'll get eeo as the respective polarities

means polarities of no of 1s and no of 2s will always remain same and different from 3s polarity..so (1s and 3s ) or (2s and 3s) can never become zero at the same time(different polarities)...so 1s and 2s can simultaneously become 0 and we'll be left with 3s only......
i believe this logic is correct..

[Govi]

My logic goes like this......or u can say i agree with sumit


Since the initial pattern of number 1 : 2 : 3 is even:evendd

considering a number, three sets of the operation will eliminate it in two steps and make it reappear in the third.

this would eventually lead to a pattern of 0 : 0 : 3


3 being the last digit remaining


Cheers

Govi...........

[DesiGuru]

Originally Posted by khanna_sumit

I believe that the answer lies in the no of these no.s being even or odd...lets call it polarity...

Yes that is a brilliant deduction...
I knew it was not the higher number of 3's that was the reason but something to do with the number properties...

Good job Khanna...

K

[Eccentric]

very well done sumit!

The original state of the system is:
{even, even, odd} i.e.the number of 1's, 2's and 3's resp.
Any operation toogles the state of each 3 parameters.
Hence, the final state of the system will be either of {odd, odd, even} or {even, even ,ood}.
If only one number remain this corres. to set {even,evn,odd) i.e. {0,0,1} a 3 remaining on the board.
Again if you play judiciously, this final state can always be achieved unless we start with number of only one type(i.e just 3's or 2's or 1's).

We can also put a count check, i.e. in start we have 85 numbers,each operation reduces that by 1 and after 84 operations we will have 1 number reamining on board and the same state we started with {even,even,odd}.

Had we started with {odd,odd,even} we would have obtatined same results.
This also goes on to show that if we start with {odd, odd, odd} or {even,even,even} we can't end up with just one number it will either be 1 number of each type or finally 2 numbers of any type