A.I.T.S Advanced level problem from Resonance

Probability Level 4

The numbers 1000 , 1001 , . . . . . , 2999 1000, 1001,....., 2999 have been written on a board. Each time, one is allowed to erase two numbers, say, a and b, and replace them by the number 1 2 m i n ( a , b ) \frac {1}{2}min(a, b) . After 1999 such operations, one obtains exactly one number c on the board, than which of them is correct

c > 1.5 c>1.5 none of these c > 1001 c>1001 c ≤ 2 c \leq2 c < 1 c<1

Atul Shivam by Atul Shivam , 23, India

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3 solutions

If we take a trivial case then it shows us the answer. If we think everyone comes and take the first pairs. I.e 1000&1001. Then second one takes 500&1001. And thus they continue. Then after 1999 operations the number left will be very less at about 0. Obviously less than 1.

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Jitender Sharma
May 12, 2019

Can anyone please tell the maximum value of c ????? 😏😏😏

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Harish Sasikumar
Nov 20, 2015

Even if we pair the numbers, so as to maximize c, we will end up in c less than 1.

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