document.write( "Question 1204981: If n is an integer between 1 and 96 (inclusive), what is the probability that n(n+1)(n+2) is divisible by 8?
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Algebra.Com's Answer #841556 by mccravyedwin(408)\"\" \"About 
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document.write( "The triples of consecutive integers allowable are (1,2,3) through (96,97,98).\r\n" );
document.write( "So there are 96 of them\r\n" );
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document.write( "Of the integers from 1 to 98, 49 are even and 49 are odd.\r\n" );
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document.write( "There are only 2 types of triples of 3 consecutive integers, the\r\n" );
document.write( "(even,odd,even)-type and the (odd,even,odd)-type.  \r\n" );
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document.write( "Case 1.  (even,odd,even)-type.  Since every other even integer is divisible by\r\n" );
document.write( "4, their product will always be divisible by 8. \r\n" );
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document.write( "These are the triples (1,2,3), (3,4,5), ..., (95,96,97). There are 48 of these.\r\n" );
document.write( "[That's because the middle numbers 2,4,...,96 are such that if you divide all \r\n" );
document.write( "of them by 2 you get 1,2,...,48.]\r\n" );
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document.write( "or \r\n" );
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document.write( "Case 2.  (odd,even,odd)-type. These will have a product divisible by 8 if and\r\n" );
document.write( "only if the middle number is divisible by 8.\r\n" );
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document.write( "These are the triples (7,8,9), (15,16,17),..., (95,96,97). There are 12 of these. \r\n" );
document.write( "[That's because the middle numbers 8,16,...,96 are such that if \r\n" );
document.write( "you divide all of them by 8, you get 1,2,...,12.]\r\n" );
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document.write( "So there are 48+12 = 60 triples of consecutive integers whose product is\r\n" );
document.write( "divisible by 8.\r\n" );
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document.write( "The desired probability is 60/96 which reduces to 5/8\r\n" );
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document.write( "Edwin
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