View All num of num See all Photos Bloomberg L.P. This employer has taken extra steps to respond to reviews and provide job seekers with accurate company information, photos, and reviews. Interested for your company?Learn More. www.bloomberg.com Employer Engaged Overview Reviews Salaries Interviews Jobs Photos Benefits 1.3k Reviews 4.1k Salaries 2.2k Interviews 738 Jobs Follow Add Review or Salary Follow Add Review or Salary Interview Question Financial Software Developer Interview New York, NY Bloomberg L.P. Given 10 cups to locate the bottle poisoned wine from a batch of normal ones, you can make any mixture of them and test your mixtures by mouses. However the density of poison in the mixture, the testing mouse will certainly die. And you must give all the mixtures of the 10 cups before the test. There is only one poisoned bottle. How many bottles of wine you can test at most? Tags: brain teaser, data structure See more , See less 8 Answer Add Tags Answer Interview Answer 5 Answers ▲ 6 ▼ 2^10You judge from the result of test. There is 2^10 kinds of possible conditions and each of them leads to a different result. So you can tell from 2^10 bottles.The mixtures are made as:1. label all the 2^10 bottle from 0-1023 in binary numbers.2. for each numbers, if any of the figures is 1, then mix this bottle into the corresponding cup. Like bottle 0000100101(2) =37(10) , figures in place 6, 3 and 1 are "1", so mix this bottle of wine into cup 1, 3, & 6.If and only if the mouses of cup 1, 3 and 6 die, the poisoned bottle will be 37. Interview Candidate on Dec 8, 2012 ▲ 4 ▼ I solved it a different way, but got the same answer -> 2^10. Start with base cases and work up.1. If you have one glass, can have at most two bottlesone poison, one not (if mouse dies, the bottle is poisoned. if mouse doesn't die, it's not poisoned and the untested one is)2. Two glasses, four bottlesone bottle untouched, one bottle in glass 1, one bottle in glass 2, one bottle in glasses 1 & 2. (similar idea: if no mice die, first bottle is poisoned. if mouse 1 dies, second bottle is poisoned. if mouse 2 dies, third bottle is poisoned. if both mice die, fourth bottle is poisoned)3. 3 glasses, 8 bottles distributed as follows:- 1 bottle untouched- 1 bottle in glass 1- 1 bottle in glass 2- 1 bottle in glass 3- 1 bottle in glasses 1 & 2- 1 bottle in glasses 2 & 3- 1 bottle in glasses 1 & 3- 1 bottle in glasses 1, 2, & 34....All the way up to 10 glasses Candidate on Dec 11, 2012 ▲ 3 ▼ seems a bit garbled.just put wine from one bottle in one cup, and give it to the mouse. if he dies, you are done. if he lives, add wine from a second bottle and give it to the mouse. repeat until the mouse is dead. the last bottle you opened is the poisoned one.hard to believe the question was actually worded that way. blair on Dec 11, 2012 ▲ 0 ▼ I also came with the same answer in a different way.if n bottles, then make mixture of n/2 bottles and test, you would know which half the poisoned bottle belongs to.Next take n/4 and n/8 and so on.so log2(n) = 10 , n = 2^10 sashi on Oct 15, 2013 ▲ 0 ▼ Notice that "you must give all the mixtures of the 10 cups before the test.", so you cannot test them one bottle by one bottle or half of bottles in one time wubi on Jan 7, 2014 Add Answers or Comments To comment on this, Sign In or Sign Up.