document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=486215>Question 486215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can you please help me? Can you give me answers to those problems and explain how to solve them? Thanks a lot!!!!<BR>\n" );
document.write( "1. The inverse of a modulo 39 is b. What is the inverse of 4a modulo 39 in terms of b?<BR>\n" );
document.write( "2. Let x and y be integers satisfying 41x + 53y = 12. Find the residue of x modulo 53.<BR>\n" );
document.write( "3. John and Ling start their new jobs on the same day. John's schedule is 4 workdays followed by 1 day off. Ling's schedule is 7 workdays followed by 2 days off. On how many days during their first year of work (365 days) do John and Ling have the same day off?<BR>\n" );
document.write( "Thank you very much!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #332405 by </B><A HREF=/tutors/aboutme.mpl?userid=chessace><B>chessace(471)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=chessace><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=chessace><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=332405\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1. b = inverse of (variable) a mod 39 means a + b = 0 mod 39.<BR>\n" );
document.write( "As usual let x = answer = inverse 4a mod 39.<BR>\n" );
document.write( "So x + 4a = 0 mod 39.<BR>\n" );
document.write( "Substract 4a+4b, which is legal because 4(a+b) = 4*0 = 0 all mod 39<BR>\n" );
document.write( "Result: x - 4b = 0 mod 39. or<BR>\n" );
document.write( "x = 4b mod 39.<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "2. 41x + 53y = 12, find x mod 53.<BR>\n" );
document.write( "53 mod 53 is 0, so the y term goes away.<BR>\n" );
document.write( "41 + 12 = 53 = 0 mod 53, so 41x = -12x mod 53<BR>\n" );
document.write( "-12x = 12 mod 53<BR>\n" );
document.write( "x = -1 mod 53<BR>\n" );
document.write( "Normalize to residue: x = 52 mod 53<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "3. The mods to use are the total cycle = work + off.<BR>\n" );
document.write( "J's days off are all those = 0 mod 5 from start 1.<BR>\n" );
document.write( "L's days off are all those = 0 or 8 mod 9 from start 1.<BR>\n" );
document.write( "So every 5*9 = 45 day interval looks the same to them.<BR>\n" );
document.write( "Within that interval, J is off 9 times which are days<BR>\n" );
document.write( " (mod 9) 5 1 6 2 7 3 8 4 0.<BR>\n" );
document.write( "You could leave out this step knowing all digits will appear once, but there is a fraction in the works, so order matters.<BR>\n" );
document.write( "Two days match every cycle, 365/45 = 8.11 cycles (no match in .11)<BR>\n" );
document.write( "So 16 days off in common that 1st year.<BR>\n" );
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