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\n" ); document.write( "Finding a solution by pure trial and error is a good way to solve the problem; but it doesn't teach you any mathematical methods.
\n" ); document.write( "There are at least a couple of ways to solve this using logical reasoning and simple mental arithmetic. Let's look at those methods first.
\n" ); document.write( "One informal method is using a \"greedy\" algorithm, in which you start by trying the largest possible number of the largest bills.
\n" ); document.write( "There is a total of 530 pesos, so the maximum number of 100-peso bills is 5. But we can't make the remaining30 pesos using five 20- or 50-peso bills.
\n" ); document.write( "If there are 4 100-peso bills, that leaves 130 pesos to be made using six 20- and 50-peso bills. But trial and error (or formal calculations) show we can't do that.
\n" ); document.write( "If there are 3 100-peso bills, that leaves 230 pesos to be made using seven 20- and 50-peso bills. Trial and error shows we CAN do that -- using 4 20-peso bills and 3 50-peso bills.
\n" ); document.write( "ANSWER: 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills = 300+150+80 = 530 pesos
\n" ); document.write( "A more sophisticated but still informal solution uses logical reasoning to quickly cut down the number of possible combinations.
\n" ); document.write( "Any combination of 50- and 100-peso bills will make a total that is a multiple of 50 pesos; the total is 530 pesos. Subtracting 20-peso bills to reduce the remaining total to a multiple of 50 pesos, we find there must be 4 20-peso bills to make 80 pesos, leaving 450 pesos to be made with the 100- and 50-peso bills.
\n" ); document.write( "Then trial and error shows the way to make the remaining 450 pesos using 6 100- and 50-peso bills is with 3 of each.
\n" ); document.write( "And again the answer is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.
\n" ); document.write( "And here is a formal algebraic solution....
\n" ); document.write( "x = # of 100-peso bills
\n" ); document.write( "y = # of 50-peso bills
\n" ); document.write( "z = # of 20-peso bills
\n" ); document.write( "[1] x+y+z = 10 (the total number of bills is 10)
\n" ); document.write( "[2] 100x+50y+20z = 530 (the total value is 530 pesos)
\n" ); document.write( "Simplify [2]: 10x+5y+2z=53
\n" ); document.write( "Eliminate z:
\n" ); document.write( "2x+2y+2z=20
\n" ); document.write( "10x+5y+2z=53
\n" ); document.write( "8x+3y=33
\n" ); document.write( "Solve that equation for one variable in terms of the other, and use the fact that x, y, and z are positive integers less than 10 to deduce the solution.
\n" ); document.write( "3y=33-8x
\n" ); document.write( "y=11-(8/3)x
\n" ); document.write( "y is an integer, and 11 is an integer, so (8/3)x must be an integer. That means x must be either 3 or 6. x=6 would make y negative, so x must be 3.
\n" ); document.write( "Then y=11-(8/3)3=11-8=3; and x=3 and y=3 means z=4.
\n" ); document.write( "And once again, using formal mathematics, the solution is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.
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