cinghal1 Posted June 3, 2007 Share Posted June 3, 2007 Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z? a. 10% b. 25% c. 50% d. 90% e. 100% PLEASE EXPLAIN YOUR ANSWER! Quote Link to comment Share on other sites More sharing options...
EEMan Posted June 3, 2007 Share Posted June 3, 2007 I'm not sure about the answer, but IMO, if 678,463 was a multiple of 6^K, 678,463 would be divisible by 6. But since 678,463 is not divisible by 6, so definitely the probability that 678,463 is not a multiple of 6^K is 100%. Are you sure that Z=6^K? or K^6? Quote Link to comment Share on other sites More sharing options...
cinghal1 Posted June 3, 2007 Author Share Posted June 3, 2007 EEMan - super explainition. I guess you are right. I just got so intimated by the text of the Q - couldnt think its so simple!! Quote Link to comment Share on other sites More sharing options...
r0m3416 Posted June 4, 2007 Share Posted June 4, 2007 Given: 6^k It means at the bare minimum 2 should be a factor (2 is factor 6 in 6^k). As the last digit in the given number 678463 is not divisible by 2 the result is 100% Quote Link to comment Share on other sites More sharing options...
sadz4u Posted February 24, 2013 Share Posted February 24, 2013 Hey man. Since 678,463 is not divisible by 6. That means the only case where it could be a multiple of z is when k (in 6^k) is 0, as that would make your z value equal to 1, and we know that anything is divisible by 1. For k to be equal to 0, our integer value which can range from 0-10 can only be 0 for 678,463 to be a mutiple of. Since there are 5 numbers in the set, there can be 5 events where our z value is 0. The total number of possible events is: (the size of our set)x(the number of available integer values) since there are 5 numbers in our set and 11 available integer values(0,1,2,3,4,5,6,7,8,9,10) the total number of possible events is : 11*5=55. (probability of our number being a multiple of z) = (the number of events where our z value is 0)/(the total number of events) = 5/55 = .091 probablility of our number NOT being a multiple z) = 1 - (probability of our number being a multiple of z) = 1 - (.091) = 0.9 the answer is therefore 90%. Quote Link to comment Share on other sites More sharing options...
Ndjamena Posted July 8, 2013 Share Posted July 8, 2013 Actually the integers range from 0-9 giving us a probability of exactly 90%. But i agree with your method. Quote Link to comment Share on other sites More sharing options...
Tristar Posted May 17, 2015 Share Posted May 17, 2015 The credit for answering this question goes to sadz4u. But, here is a small correction. Actually, we have 10 first non-negative integers which are the numbers from 0 to 9. Then, if we consider the probability of our required number NOT being a multiple of Z then there are 9 such integers out of 10. Quote Link to comment Share on other sites More sharing options...
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