<PRE><B>Question 1187485</B>
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Finding a solution by pure trial and error is a good way to solve the problem; but it doesn&#39;t teach you any mathematical methods.&lt;br&gt;
There are at least a couple of ways to solve this using logical reasoning and simple mental arithmetic.  Let&#39;s look at those methods first.&lt;br&gt;
One informal method is using a &quot;greedy&quot; algorithm, in which you start by trying the largest possible number of the largest bills.&lt;br&gt;
There is a total of 530 pesos, so the maximum number of 100-peso bills is 5.  But we can&#39;t make the remaining30 pesos using five 20- or 50-peso bills.
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&#39;t do that.
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.&lt;br&gt;
ANSWER: 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills = 300+150+80 = 530 pesos&lt;br&gt;
A more sophisticated but still informal solution uses logical reasoning to quickly cut down the number of possible combinations.&lt;br&gt;
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.&lt;br&gt;
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.&lt;br&gt;
And again the answer is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.&lt;br&gt;
And here is a formal algebraic solution....&lt;br&gt;
x = # of 100-peso bills
y = # of 50-peso bills
z = # of 20-peso bills&lt;br&gt;
[1] x+y+z = 10  (the total number of bills is 10)
[2] 100x+50y+20z = 530  (the total value is 530 pesos)&lt;br&gt;
Simplify [2]: 10x+5y+2z=53&lt;br&gt;
Eliminate z:&lt;br&gt;
2x+2y+2z=20
10x+5y+2z=53
8x+3y=33&lt;br&gt;
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.&lt;br&gt;
3y=33-8x
y=11-(8/3)x&lt;br&gt;
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.&lt;br&gt;
Then y=11-(8/3)3=11-8=3; and x=3 and y=3 means z=4.&lt;br&gt;
And once again, using formal mathematics, the solution is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.&lt;br&gt;
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