document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=933275>Question 933275</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help! I've tried several times to figure out this question and it just flusters me. \"What are the sets of consecutive numbers that add up to 315?\" I have found one and it is 104+105+106=315 and there are 11 total ways. I understand you are busy but i would really appriciate if you could help me out. Thank you.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #566718 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=566718\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Also 157+158.\r<BR>\n" );
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document.write( "More generally, suppose <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large a + (a+1) + (a+2) + \ldots + (a+d) = 315\"> where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large d \ge 1\">. The LHS equals <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large a(d+1) + (1+2+\ldots + d) = a(d+1) + \frac{d(d+1)}{2} = 315\"> so <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large (d+1)(a + \frac{d}{2}) = 315\">.\r<BR>\n" );
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document.write( "Multiply both sides by 2 to get <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large (d+1)(2a+d) = 630\">. Note that d+1 must necessarily be a factor of 630 that is at least 2. Once we fix d, we can uniquely determine a (note that d odd --> 2a+d odd, d even --> 2a+d even). The factors of 630 are: 1,2,3,5,6,7,9,10,14,15,18,21,30,35,42,45,63,70,90,105,126,210,315,630 (24 divisors), so d is 1 less than a factor of 630. The following list the values of a for each value of d:\r<BR>\n" );
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document.write( "d: a<BR>\n" );
document.write( "0 315.0<BR>\n" );
document.write( "1 157.0<BR>\n" );
document.write( "2 104.0<BR>\n" );
document.write( "4 61.0<BR>\n" );
document.write( "5 50.0<BR>\n" );
document.write( "6 42.0<BR>\n" );
document.write( "8 31.0<BR>\n" );
document.write( "9 27.0<BR>\n" );
document.write( "13 16.0<BR>\n" );
document.write( "14 14.0<BR>\n" );
document.write( "17 9.0<BR>\n" );
document.write( "20 5.0<BR>\n" );
document.write( "29 -4.0<BR>\n" );
document.write( "34 -8.0<BR>\n" );
document.write( "41 -13.0<BR>\n" );
document.write( "44 -15.0<BR>\n" );
document.write( "62 -26.0<BR>\n" );
document.write( "69 -30.0<BR>\n" );
document.write( "89 -41.0<BR>\n" );
document.write( "104 -49.0<BR>\n" );
document.write( "125 -60.0<BR>\n" );
document.write( "209 -103.0<BR>\n" );
document.write( "314 -156.0<BR>\n" );
document.write( "629 -314.0\r<BR>\n" );
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document.write( "For example, if d = 5, we have 50+51+...+(50+5) = 50+51+52+53+54+55+56 = 315. If d = 629, we have -314 + (-313) + ... + 315 = 315.\r<BR>\n" );
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document.write( "So in fact there are 24 possible ways to find consecutive numbers adding to 315 (including d = 0), but if we only include positive integers, and exclude d = 0, there are 11 ways total (d = 1, 2, ..., 17, 20). </SPAN>\n" );
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