Puzzle #5: Save the king

Problem: There are 500 barrels of wine and exactly one of them is poisoned. The king wants to identify the poisoned barrel but he is ready to sacrifice only 4 prisoners. They have 4 days to find the poisoned barrel. The poison has the property that after consuming it one can die anytime in the next 24hrs. Give the strategy to find the poisoned barrel.

Source: Shahbaz  also CSE blog

Puzzle #4: Stop the roll

Problem: You are in a dicey situation. You friend gave you a dice and asked you to keep rolling till you get a sum of 100 or more. Now,you have to tell the most probable number at which you are going to stop.

Solution: You will stop at or before 105. Now, trick is to think backwards. You can get 105, only if you ever reach 99. Similarly, you can get 104, only from 99 or 98. You can get to 100, from maximum previous sums i.e. 94, 95, 96, 97, 98 and 99. Therefore, 100 is the most probable stopping point.  P(105) = P(105|99).P(99) = P(99)/6 . Similiarly, P(104) = (P(99)+P(98))/6, ...P(100)  = (P(94)+...+P(99))/6

Source: Rishab

Maths puzzle #3- XOR magic

Problem: You are given a set of n numbers, C = { c_1, c_2, ..., c_n}and you have to find the minimum size of subset of C from which you can create the given n numbers by only doing xor operations on the elements of the subset.

Solution: One starts by thinking about the properties of XOR. There are only a few properties that I know, like x + y + x = y ( I'll represent XOR by + through out the post). So, I have to find a subset V = { v_1, v_2,...,v_k} such that a_i1*v_1+a_i2*v_2+...+a_ik*v_k = c_i, for all elements of C and a_ij are either 0 or 1. This looks similat to c_i being linear combination of V. So, we check if there is a vector space with XOR as the addition operation and voilà, if a euclidean vector space is over the field Z_2, then addition operation is x+y % 2 which is the definition of XOR. So, now all we need to find is the rank of matrix [c_1 c_2 ...c_n] . To find the rank reduced matrix by elementary operations (gauss elimination)

Source: Rishab Vaid  who read it perhaps on Topcoder

Maths puzzle #2 - Divisibility by 100

Given any 51 integers, prove that there exist two integers a, b such that a^2 - b^2  is divisible by 100

Solution: Start thinking by why only 51? If we divide any number by 50 it will give 50 possible remainders, which means there are two integers a, b which give the same remainder. So, a = 50q1+r and b = 50q2+r where 0<= r < 50.

a^2 - b^2 = (a-b)*(a+b) = 50(q1-q2) * 2(25(q1+q2)+r), hence divisibility by 100

Source: Manish asked me this problem, who was in turn asked by Rustam, who read it in Mathematical circles: the russian experience