document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=401152>Question 401152</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A child has $6.50 in dimes and quarters. Determine the number of possible combinations of coins he/she can have. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #283845 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=283845\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "The total amount for any odd number of quarters ends in a 5 in the cents column.  Since $6.50 ends in a zero in the cents column and our young stalwart has no nickels, she must have an even number of quarters.\r<BR>\n" );
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document.write( "The statement \"...$6.50 in dimes and quarters.\" implies that the number both the number of dimes and the number of quarters must be greater than zero.\r<BR>\n" );
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document.write( "Since 26 quarters is exactly $6.50, 24 quarters is the most she could have, and that would have to be paired with 5 dimes.  (24 X 25 = 600 and 5 X 10 = 50)\r<BR>\n" );
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document.write( "Since 65 dimes is exactly $6.50 and the fewest quarters possible is 2, the most dimes possible is 60 paired with the 2 quarters.  (60 X 10 = 600 and 2 X 25 = 50)\r<BR>\n" );
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document.write( "So\r<BR>\n" );
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document.write( "24Q + 5D<BR>\n" );
document.write( "22Q + 10D<BR>\n" );
document.write( "20Q + 15D\r<BR>\n" );
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document.write( "and so on until you get to\r<BR>\n" );
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document.write( "6Q + 50D<BR>\n" );
document.write( "4Q + 55D<BR>\n" );
document.write( "2Q + 60D\r<BR>\n" );
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document.write( "Total number of combinations, either 24 divided by 2 or 60 divided by 5.  If it turns out that the \"both denominations greater than zero\" assumption is incorrect, then add two to your total count.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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