document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1143478>Question 1143478</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the smallest positive integer which, when divided by each of 2, 3, 4, 5, 6<BR>\n" );
document.write( "and 7, will give in each case a remainder that is one less than the divisor? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #764283 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</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=764283\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "You are looking for the number N which, when divided by any of 2, 3, 4, 5, 6, or 7, leaves a remainder that is 1 less than the divisor.<br><BR>\n" );
document.write( "With those conditions, the number N+1 will be divisible by each of those divisors.<br><BR>\n" );
document.write( "So the number you are looking for is the smallest positive integer that is 1 less than the least common multiple of 2, 3, 4, 5, 6, and 7.<br><BR>\n" );
document.write( "The LCM of those divisors is 2*2*3*5*7 = 420; the number you are looking for is 419.<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( "419/2 = 209 remainder 1<BR>\n" );
document.write( "419/3 = 139 remainder 2<BR>\n" );
document.write( "419/4 = 104 remainder 3<BR>\n" );
document.write( "419/5 = 83 remainder 4<BR>\n" );
document.write( "419/6 = 69 remainder 5<BR>\n" );
document.write( "419/7 = 59 remainder 6 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );