document.write( "Question 751645: What are all the possible combinations of coins( using pennies, nickels, dimes and quarters) that equal 43 cents. \n" ); document.write( "
Algebra.Com's Answer #457320 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "The only way to solve this is to find a logical way to list them.\r
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\n" ); document.write( "\n" ); document.write( "Two quarters is 50 cents, so it is only possible to have either one quarter or zero quarters.\r
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\n" ); document.write( "\n" ); document.write( "Start with 1 quarter. That leaves you with 18 cents to make up. So, given that you have 1 quarter, the most dimes you can have is 1, leaving you 8 cents to make up, so you can have, at most, 1 nickel, and then 3 pennies.\r
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\n" ); document.write( "\n" ); document.write( "Now take away the nickel and add 5 pennies, so 1 Q, 1 D, 0 N, 8 P.\r
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\n" ); document.write( "\n" ); document.write( "Take away the dime, so:\r
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\n" ); document.write( "\n" ); document.write( "1 Q, 0 D, 3 N, 3 P, or
\n" ); document.write( "1 Q, 0 D, 2 N, 8 P, or
\n" ); document.write( "1 Q, 0 D, 1 N, 13 P, or
\n" ); document.write( "1 Q, 0 D, 0 N, 18 P\r
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\n" ); document.write( "\n" ); document.write( "Take away the quarter, and now you can have up to 4 dimes:\r
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\n" ); document.write( "\n" ); document.write( "0 Q, 4 D, 0 N, 3 P\r
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\n" ); document.write( "\n" ); document.write( "And so on until you get down to 0 Q, 0 D, 0 N, 43 P. You can follow the pattern and then count the possibilities for yourself.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "Egw to Beta kai to Sigma
\n" ); document.write( "My calculator said it, I believe it, that settles it
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