BC2 / NC2 = 0.5

B(B-1)*2 = N*(N-1)

One of N or N-1 is a multiple of 4

N=4,5,8,9,12,13…

N=4, B=3 fits the bill on first try ðŸ™‚

]]>I asked my son Anahat to attempt this question. He mentioned that there are three sets of numbers which need to be ruled out in the very beginning. 1 through 9, 50, and 91 though 99. That leaves us with two sets of numbers. 10-49 and 51-99. Now, in order to avoid 100, we choose from all odd numbers from the first set (10 – 49) and all even numbers from the set (51-99) or reverse. This leads to a total of 1 (number 50) + 9 (from 91 through 99) + 20 (odd selection )+ 20 (even selection). This adds up to 50 numbers. Any additional number chosen will enable an even+even or an odd+odd operation which can provide 100 as a solution. Therefore we can say that more than 50 numbers on the blackboard without getting 100 in total is not possible.

– Prakhar ]]>

Is it possible to have 50 numbers from [-40, 49] such that sum of no two numbers is 0.

[41,49] would not matter here

So, Is it possible to have 41 numbers from [-40, 40] such that sum of no two numbers is 0.

Not possible.

]]>When 6 stars are placed in 4 rows, either at least one row has at least 3 stars or at least two has two stars each. So you can always find 2 rows with 4 stars. Pick these two rows and remove them. Now 2 stars remain and we can remove two columns. So, we can always remove all stars by removing two rows and two columns. ]]>