document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4611>Question 4611</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Is 2^n3^2n-1 always divisible by 17? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2176 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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=2176\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Yes, proof as below.\r<BR>\n" );
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document.write( " 2^n3^2n -1  = (2*3^2)^n - 1 = (18^n) - 1\r<BR>\n" );
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document.write( " Use a^n-b^n = (a-b)(a^(n-1) + a^(n-2)b+...+ ab^(n-2)+ b^(n-1))<BR>\n" );
document.write( " Now a= 18, b = 1,<BR>\n" );
document.write( " we have (18^n) - 1 = (18-1)(18^17+18^16+...+18+1) <BR>\n" );
document.write( "  = 17*(18^17+18^16+...+18+1)\r<BR>\n" );
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document.write( " This shows 17 is a divisor of  2^n3^2n -1<BR>\n" );
document.write( " Or proved by Mathematical Induction.\r<BR>\n" );
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document.write( " Kenny </SPAN>\n" );
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