document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1152784>Question 1152784</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Roger	has	27	standard	U.S.	coins	worth	a	total	of	$3.65.	No	coin	is	worth	more	than	30¢	or	less	<BR>\n" );
document.write( "than	3¢.	How	many	dimes	could	Roger	have?	(NOTE:	There	are	several	possible	answers,	but	<BR>\n" );
document.write( "you	only	need	to	give	TWO	correct	answers.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #774877 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=774877\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "List out the various coins<BR>\n" );
document.write( "{penny, nickel, dime, quarter, half dollar, silver dollar}<BR>\n" );
document.write( "We are told that no coin is less than 3 cents, so we can cross off the penny<BR>\n" );
document.write( "{<s>penny</s>, nickel, dime, quarter, half dollar, silver dollar}<BR>\n" );
document.write( "leaving us with this set<BR>\n" );
document.write( "{nickel, dime, quarter, half dollar, silver dollar}\r<BR>\n" );
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document.write( "and we are also told that no coin is worth more than 30 cents, so we can cross off the half dollar and silver dollar coins<BR>\n" );
document.write( "{nickel, dime, quarter, <s>half dollar</s>, <s>silver dollar</s>}<BR>\n" );
document.write( "leaving us with<BR>\n" );
document.write( "{nickel, dime, quarter}\r<BR>\n" );
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document.write( "Therefore, the coins we're dealing with are: {nickel, dime, quarter}\r<BR>\n" );
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document.write( "n = number of nickels<BR>\n" );
document.write( "d = number of dimes<BR>\n" );
document.write( "q = number of quarters\r<BR>\n" );
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document.write( "n+d+q = 27 because we have 27 coins<BR>\n" );
document.write( "Solve for q to get q = 27-n-d\r<BR>\n" );
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document.write( "5n = value, in cents, of all the nickels <BR>\n" );
document.write( "10d = value, in cents, of all the dimes<BR>\n" );
document.write( "25q = value, in cents, of all the quarters<BR>\n" );
document.write( "5n+10d+25q = total value of all the coins<BR>\n" );
document.write( "total value = 365 cents = 3.65 dollars<BR>\n" );
document.write( "5n+10d+25q = 365\r<BR>\n" );
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document.write( "5n+10d+25q = 365<BR>\n" );
document.write( "5n+10d+25(27-n-d) = 365 ... plug in q = 27-n-d<BR>\n" );
document.write( "5n+10d+25(27)+25(-n)+25(-d) = 365 ... distribute<BR>\n" );
document.write( "5n+10d+675-25n-25d = 365<BR>\n" );
document.write( "-20n-15d+675 = 365<BR>\n" );
document.write( "-20n-15d+675-675 = 365-675 ... subtract 675 from both sides<BR>\n" );
document.write( "-20n-15d = -310 <BR>\n" );
document.write( "-5(4n+3d) = -310 <BR>\n" );
document.write( "4n+3d = 62 .... divide both sides by -5\r<BR>\n" );
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document.write( "Now solve for n<BR>\n" );
document.write( "4n+3d = 62<BR>\n" );
document.write( "4n = -3d+62<BR>\n" );
document.write( "n = (-3d+62)/4<BR>\n" );
document.write( "n = (-3d+2+60)/4<BR>\n" );
document.write( "n = (-3d+2)/4+60/4<BR>\n" );
document.write( "n = (-3d+2)/4+15<BR>\n" );
document.write( "Through trial and error, you'll find that d = 2 makes the expression (-3d+2)/4 result in an integer<BR>\n" );
document.write( "n = (-3d+2)/4+15<BR>\n" );
document.write( "n = (-3*2+2)/4+15<BR>\n" );
document.write( "n = (-6+2)/4+15<BR>\n" );
document.write( "n = -4/4+15<BR>\n" );
document.write( "n = -1+15<BR>\n" );
document.write( "n = 14<BR>\n" );
document.write( "So we have d = 2 dimes pair up with n = 14 nickels.\r<BR>\n" );
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document.write( "If d = 2 and n = 14, then,<BR>\n" );
document.write( "q = 27-n-d<BR>\n" );
document.write( "q = 27-2-14<BR>\n" );
document.write( "q = 11\r<BR>\n" );
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document.write( "Now increase d = 2 to d = 6. This is an increase of +4, which is from the denominator of 4 in (-3d+2)/4\r<BR>\n" );
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document.write( "We have,<BR>\n" );
document.write( "n = (-3d+2)/4+15<BR>\n" );
document.write( "n = (-3*6+2)/4+15<BR>\n" );
document.write( "n = (-18+2)/4+15<BR>\n" );
document.write( "n = -16/4+15<BR>\n" );
document.write( "n = -4+15<BR>\n" );
document.write( "n = 11\r<BR>\n" );
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document.write( "Now use n = 11 and d = 6 to get,<BR>\n" );
document.write( "q = 27-n-d<BR>\n" );
document.write( "q = 27-11-6<BR>\n" );
document.write( "q = 10\r<BR>\n" );
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document.write( "Two solutions are<BR>\n" );
document.write( "(n,d,q) = (14,2,11)<BR>\n" );
document.write( "and<BR>\n" );
document.write( "(n,d,q) = (11,6,10)\r<BR>\n" );
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document.write( "For the solution (n,d,q) = (14,2,11), we can say<BR>\n" );
document.write( "5n = 5*14 = 70 cents is from the nickels only<BR>\n" );
document.write( "10d = 10*2 = 20 cents is from the dimes only<BR>\n" );
document.write( "25q = 25*11 = 275 cents is from the quarters only<BR>\n" );
document.write( "5n+10d+25q = 70+20+275 = 365 cents = $3.65 is the total combined value<BR>\n" );
document.write( "This confirms the solution (n,d,q) = (14,2,11)\r<BR>\n" );
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document.write( "For the solution (n,d,q) = (11,6,10), we can say<BR>\n" );
document.write( "5n = 5*11 = 55 cents is from the nickels only<BR>\n" );
document.write( "10d = 10*6 = 60 cents is from the dimes only<BR>\n" );
document.write( "25q = 25*10 = 250 cents is from the quarters only<BR>\n" );
document.write( "5n+10d+25q = 55+60+250 = 365 cents = $3.65 is the total combined value<BR>\n" );
document.write( "This confirms the solution (n,d,q) = (11,6,10)\r<BR>\n" );
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document.write( "There are 5 total ways to have nickels, dimes, and quarters combine to $3.65 such that we have 27 coins total<BR>\n" );
document.write( "<TABLE  border = \"1\" cellpadding = \"5\">\n" );
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document.write( "<tr><TD >Case</TD><TD >Nickels</TD><TD >Dimes</TD><TD >Quarters</TD></tr>\r\n" );
document.write( "<tr><TD >A</TD><TD >2</TD><TD >18</TD><TD >7</TD></tr>\r\n" );
document.write( "<tr><TD >B</TD><TD >5</TD><TD >14</TD><TD >8</TD></tr>\r\n" );
document.write( "<tr><TD >C</TD><TD >8</TD><TD >10</TD><TD >9</TD></tr>\r\n" );
document.write( "<tr><TD >D</TD><TD >11</TD><TD ><font color=red>6</font></TD><TD >10</TD></tr>\r\n" );
document.write( "<tr><TD >E</TD><TD >14</TD><TD ><font color=red>2</font></TD><TD >11</TD></tr>\r\n" );
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document.write( "Each row adds to 27.<BR>\n" );
document.write( "The solutions we found earlier are in cases D and E of this table.\r<BR>\n" );
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document.write( "So Roger could have <font color=red>6 dimes</font> (case D) or he could have <font color=red>2 dimes</font> (case E).<BR>\n" );
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