document.write( "Question 388792: Two fair dice, each with faces numbered from 1 thru 6, are rolled at the same time. Each dice has five exposed faces, which are summed. Express as a common fraction the probability that the least common multiple of the two sums of the exposed faces is a multiple of six. \n" ); document.write( "
Algebra.Com's Answer #275276 by Edwin McCravy(20060)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "The sum of all 6 sides of a die is 1+2+3+4+5+6 = 21\r\n" ); document.write( "\r\n" ); document.write( "So the 'Sum of all the Exposed Sides' (which I'll shorten to SES) of a die is\r\n" ); document.write( "21 minus whichever side is on the bottom, unexposed. So,\r\n" ); document.write( "\r\n" ); document.write( "The SES of a rolled die is 21-1 = 20 if 1 is on the bottom\r\n" ); document.write( "The SES of a rolled die is 21-2 = 19 if 2 is on the bottom\r\n" ); document.write( "The SES of a rolled die is 21-3 = 18 if 3 is on the bottom\r\n" ); document.write( "The SES of a rolled die is 21-4 = 17 if 4 is on the bottom\r\n" ); document.write( "The SES of a rolled die is 21-5 = 16 if 5 is on the bottom\r\n" ); document.write( "The SES of a rolled die is 21-6 = 15 if 6 is on the bottom\r\n" ); document.write( "\r\n" ); document.write( "In order that the LCM of the SES's of the dice be a multiple of 6, then either:\r\n" ); document.write( "\r\n" ); document.write( "Case 1. The SES of one of them must have a factor of 6 \r\n" ); document.write( "\r\n" ); document.write( "or else\r\n" ); document.write( "\r\n" ); document.write( "Case 2. The SES of one of them must have a factor of 3, not 6, and the SES of\r\n" ); document.write( "the other must have a factor of 2. \r\n" ); document.write( "\r\n" ); document.write( "Case 1:\r\n" ); document.write( "Only one of the possible SES's contains a factor of 6, and that is 18, which\r\n" ); document.write( "occurs when 3 is on the bottom. 18 is divisible by 6. So If 3 is on the bottom\r\n" ); document.write( "then the SES is 18. Therefore in this case, it doesn't matter what the SES of\r\n" ); document.write( "the other die is, the LCM will be divisible by 18, and thus divisible by 6. So\r\n" ); document.write( "all rolls when one of the dice has a 3 on the bottom will have an SES which is\r\n" ); document.write( "divisible by 6. There are 11 such rolls that have these pairs of sides on the\r\n" ); document.write( "bottom:\r\n" ); document.write( "\r\n" ); document.write( "(1,3), (2,3),(3,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6),\r\n" ); document.write( "\r\n" ); document.write( " \r\n" ); document.write( "Case 2:\r\n" ); document.write( "If neither SES is 18, then one of the SES's must be 15. That's because 15 is\r\n" ); document.write( "the only possible SES besides 18 which has a factor of 3, and that occurs when\r\n" ); document.write( "a 6 is on the bottom. Since an SES of 18 is eliminated in this case, there\r\n" ); document.write( "are only 2 SES's remaining with a factor of 2, namely, 20 and 16. That's when\r\n" ); document.write( "a 1 or 5 is on the bottom. There are 4 such rolls that have these pairs of\r\n" ); document.write( "sides on the bottom. \r\n" ); document.write( "\r\n" ); document.write( "(1,6), (5,6), (6,1), (6,5) \r\n" ); document.write( "\r\n" ); document.write( "So there are 11+4 = 15 ways the dice can be rolled and the LCM of the SES's be\r\n" ); document.write( "divisible by 6.\r\n" ); document.write( "\r\n" ); document.write( "There are 36 possible rolls, so the desired probability is\n" ); document.write( " |