document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1208429>Question 1208429</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the last two digits of the number 3^123 + 7^123 + 9^123. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #846837 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=846837\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Look for the repeating pattern of the last two digits of 3, 7, and 9 to increasing powers.<br><BR>\n" );
document.write( "3: 03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01, 03...<br><BR>\n" );
document.write( "That pattern repeats with a cycle length of 20. 123 mod 20 = 3, so the last two digits of 3^123 is the 3rd number in the pattern: 27<br><BR>\n" );
document.write( "7: 07, 49, 43, 01, 07...<br><BR>\n" );
document.write( "That pattern repeats with a cycle length of 4. 123 mod 4 = 3, so the last two digits of 7^123 is the 3rd number in the pattern: 43<br><BR>\n" );
document.write( "9: 09, 81, 29, 61, 49, 41, 69, 21, 89, 01, 09 ...<br><BR>\n" );
document.write( "That pattern repeats with a cycle length of 10. 123 mod 10 = 3, so the last two digits of 9^123 is the 3rd number in the pattern: 29<br><BR>\n" );
document.write( "27+43+29 = 99<br><BR>\n" );
document.write( "ANSWER: 99<br><BR>\n" );
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