document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204981>Question 1204981</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If n is an integer between 1 and 96 (inclusive), what is the probability that n(n+1)(n+2) is divisible by 8? <BR>\n" );
document.write( "a.1/4 b.1/2 c.5/8 d.3/4 e.7/8 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #841560 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=841560\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Answer:   <font color=red size=4>5/8  (choice C)</font>\r<BR>\n" );
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document.write( "Explanation\r<BR>\n" );
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document.write( "Even though the other tutors offer much more efficient solutions, I'll post mine anyway. <BR>\n" );
document.write( "It follows the same idea as tutor ikleyn, but I show a bit more detail.\r<BR>\n" );
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document.write( "When dividing by 8, there are 8 possible remainders: 0 through 7.<BR>\n" );
document.write( "\"remainder 0\" is the same as \"remainder 8\".\r<BR>\n" );
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document.write( "The class of numbers that give remainder 0 are: 8, 16, 24, 32, ...<BR>\n" );
document.write( "i.e. the multiples of 8<BR>\n" );
document.write( "Let k be an integer<BR>\n" );
document.write( "n = 8k = some multiple of 8<BR>\n" );
document.write( "It's fairly obvious that n(n+1)(n+2) is a multiple of 8 when n = 8k<BR>\n" );
document.write( "If you aren't convinced, then replace n with 8k to see that:<BR>\n" );
document.write( "n(n+1)(n+2) = 8k(8k+1)(8k+2) = 8*( k(8k+1)(8k+2) ) = 8*(some integer)\r<BR>\n" );
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document.write( "Now onto the \"remainder 1\" class of numbers (1, 9, 17, 25, ...)<BR>\n" );
document.write( "Those are of the form n = 8k+1. They are 1 more than a multiple of 8.<BR>\n" );
document.write( "Plug that into n(n+1)(n+2) to get (8k+1)(8k+2)(8k+3)<BR>\n" );
document.write( "Expand that out to get the cubic 512k^3 + 384k^2 + 88k + 6<BR>\n" );
document.write( "Then notice how the first three terms we can factor out 8, so we could rewrite that as<BR>\n" );
document.write( "8(64k^3 + 48k^2 + 11k) + 6 = 8(some integer) + 6<BR>\n" );
document.write( "Values of n in the \"remainder 1\" class lead to n(n+1)(n+2) having remainder 6.\r<BR>\n" );
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document.write( "Example:<BR>\n" );
document.write( "n = 9<BR>\n" );
document.write( "n(n+1)(n+2) = 9*10*11 = 990<BR>\n" );
document.write( "990/8 = 123 remainder 6<BR>\n" );
document.write( "Another example:<BR>\n" );
document.write( "n = 17<BR>\n" );
document.write( "n(n+1)(n+2) = 17*18*19 = 5814<BR>\n" );
document.write( "5814/8 = 726 remainder 6\r<BR>\n" );
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document.write( "So we can rule out all of the \"remainder 1\" class numbers.\r<BR>\n" );
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document.write( "For the \"remainder 2\" class, we could follow similar steps to remainder 1.<BR>\n" );
document.write( "But this time I'll use modular arithmetic as a shortcut. <BR>\n" );
document.write( "If you aren't familiar with this notation, then please review online resources. <BR>\n" );
document.write( "You could get away with doing this problem without resorting to modular arithmetic (by using the polynomial method shown above), but the polynomial method is cumbersome.\r<BR>\n" );
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document.write( "Here's what the remainder 2 case looks like<BR>\n" );
document.write( "n = 8k+2<BR>\n" );
document.write( "n(n+1)(n+2)<BR>\n" );
document.write( "= (8k+2)(8k+2+1)(8k+2+2)<BR>\n" );
document.write( "= (8k+2)(8k+3)(8k+4)<BR>\n" );
document.write( "= 2*3*4 (mod 8)<BR>\n" );
document.write( "= 24 (mod 8)<BR>\n" );
document.write( "= 0 (mod 8)<BR>\n" );
document.write( "We get remainder 0 at the end, proving that n = 8k+2 leads to n(n+1)(n+2) to be a multiple of 8.<BR>\n" );
document.write( "Therefore numbers like n = 2, n = 10, n = 18, n = 26, etc will make n(n+1)(n+2) to be a multiple of 8.\r<BR>\n" );
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document.write( "Example:<BR>\n" );
document.write( "n = 10<BR>\n" );
document.write( "n(n+1)(n+2) = 10*11*12 = 1320<BR>\n" );
document.write( "1320/8 = 165 remainder 0 = 165<BR>\n" );
document.write( "1320 is a multiple of 8, so n(n+1)(n+2) is a multiple of 8 when n = 10.\r<BR>\n" );
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document.write( "I'll go through the other remainder classes fairly quickly using the mod notation<BR>\n" );
document.write( "n = 8k+3<BR>\n" );
document.write( "n(n+1)(n+2)<BR>\n" );
document.write( "= (8k+3)(8k+3+1)(8k+3+2)<BR>\n" );
document.write( "= (8k+3)(8k+4)(8k+5)<BR>\n" );
document.write( "= 3*4*5 (mod 8)<BR>\n" );
document.write( "= 60 (mod 8)<BR>\n" );
document.write( "= 4 (mod 8)<BR>\n" );
document.write( "We don't get remainder 0, so we rule out the \"remainder 3\" class of numbers (eg: 3, 11, 19, ...)\r<BR>\n" );
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document.write( "n = 8k+4<BR>\n" );
document.write( "n(n+1)(n+2)<BR>\n" );
document.write( "= (8k+4)(8k+4+1)(8k+4+2)<BR>\n" );
document.write( "= (8k+4)(8k+5)(8k+6)<BR>\n" );
document.write( "= 4*5*6 (mod 8)<BR>\n" );
document.write( "= 120 (mod 8)<BR>\n" );
document.write( "= 8*15 (mod 8)<BR>\n" );
document.write( "= 0 (mod 8)<BR>\n" );
document.write( "We do get remainder 0, which means any value of n part of the \"remainder 4\" group will make n(n+1)(n+2) a multiple of 8.<BR>\n" );
document.write( "Some example values are: n = 4, n = 12, n = 20, n = 28, ...\r<BR>\n" );
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document.write( "n = 8k+5<BR>\n" );
document.write( "n(n+1)(n+2)<BR>\n" );
document.write( "= (8k+5)(8k+5+1)(8k+5+2)<BR>\n" );
document.write( "= (8k+5)(8k+6)(8k+7)<BR>\n" );
document.write( "= 5*6*7 (mod 8)<BR>\n" );
document.write( "= 5*(-2)*(-1) (mod 8)<BR>\n" );
document.write( "= 10 (mod 8)<BR>\n" );
document.write( "= 2 (mod 8)<BR>\n" );
document.write( "The nonzero remainder rules out everything in the \"remainder 5\" group.<BR>\n" );
document.write( "Some values in this group are 5, 13, 21, ...\r<BR>\n" );
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document.write( "Cases n = 8k+6 and n = 8k+7 are handled by n = 8k-2 and n = 8k-1 respectively<BR>\n" );
document.write( "But notice how if n = 8k-2 then the n+2 is a multiple of 8<BR>\n" );
document.write( "Similarly, n = 8k-1 leads to n+1 being a multiple of 8<BR>\n" );
document.write( "Therefore, cases n = 8k+6 and n = 8k+7 are guaranteed to lead n(n+1)(n+2) as a multiple of 8.\r<BR>\n" );
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document.write( "------------------\r<BR>\n" );
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document.write( "There's a lot to take in. <BR>\n" );
document.write( "But the key summary is that we get n(n+1)(n+2) as a multiple of 8 for these remainder classes: 2, 4, 6, 7, 8<BR>\n" );
document.write( "There are 5 remainder classes of values that will give us a multiple of 8.<BR>\n" );
document.write( "This is out of 8 remainder classes.\r<BR>\n" );
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document.write( "Ultimately this leads to the <font color=red size=4>final answer 5/8</font><BR>\n" );
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