Another old trick


There are 6 slots filled with 1 , 0 in the above order.Now you can operate on any two adjacent sectors and increase the values of both by 1 at a time.you can change the values of two adjacent sectors only at a time.Is it possible that after a sequence of operations al the sectors possess the same value ?

11 Answers

1
Aditya ·

I think not possible!

62
Lokesh Verma ·

it is not possible but why?

1
Akand ·

cos he has tried every possible way.............and ya even im not gettin it.

BUT i kno ther has to be sum gud logic and proof for not gettin it

1
Akand ·

wel......I got all d values same except one.....

1
Aditya ·

sry bhaiyya....it's just intuition[2].....n ya i tried sum values too as Akand did.

62
Lokesh Verma ·

this is a very good question...
In fact it has to do with this..

marks the circle sectors as 1, 2, 3, 4, 5, 6

the sum of the even sections will be 0 and odd is 2

whenever you add 2 numbers they get added to both the odd sector and the even sector

so the difference remains same...

62
Lokesh Verma ·

I know i have not explained it well..but i think u must have got the point

1
Aditya ·

Thanks bhaiyya...!!!

341
Hari Shankar ·

Alternately, if you number the sectors as 1,2,3,4,5, and 6 , and consider the function

S = |x_1-x_2| + |x_2-x_3|+|x_3-x_4|+|x_4-x_5| + |x_5-x_6|+|x_6-x_1|

you will see that under the conditions of the problem, every time we perform the given operation, S increases by 1.

So we can never make S = 0 which is what it would be if all the numbers became equal

1
chemistry organic ·

the zero between two 1s is creating the problem otherwise it would have been possible

1
Aditya ·

thanks prophet sir

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