document.write( "Question 127819: Daddy Warbucks always carries a specific number of $100 and $500 bills for impulse purchases and $1 bills for tips. If he has 500 bills in his briefcase and they total $50000, how many of each denomination does he carry?\r
\n" ); document.write( "\n" ); document.write( "Please help me, I do not know how to set this up to solve. Thankyou
\n" ); document.write( "My question is not from a textbook, but it is from my teacher in my Math 102, Intermediate Algebra class.
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Algebra.Com's Answer #93616 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
I used h for hundreds, f for five hundreds, and s for singles.\r
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\n" ); document.write( "\n" ); document.write( "So \"h%2Bf%2Bs=500\".\r
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\n" ); document.write( "\n" ); document.write( "Hundreds are worth $100, so the value of the h hundreds is \"100h\", likewise the value of the f five hundreds is \"500f\", and the value of the s singles is just \"s\", and we know that:\r
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\n" ); document.write( "\n" ); document.write( "\"100h%2B500f%2Bs=50000\"\r
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\n" ); document.write( "\n" ); document.write( "But now what? We have three unknowns but only two equations.\r
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\n" ); document.write( "\n" ); document.write( "The secret lies in realizing that the answers must be positive integers.\r
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\n" ); document.write( "\n" ); document.write( "Let's rearrange the equations a little:\r
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\n" ); document.write( "\n" ); document.write( "\"h%2Bs=500-f\"\r
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\n" ); document.write( "\n" ); document.write( "\"100h%2Bs=50000-500f\"\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first equation by -1:
\n" ); document.write( "\"-h-s=-500%2Bf\"\r
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\n" ); document.write( "\n" ); document.write( "And add it term-by-term to the second equation:
\n" ); document.write( "\"99h=49500-499f\"\r
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\n" ); document.write( "\n" ); document.write( "Then divide by 99:
\n" ); document.write( "\"h=%2849500-499f%29%2F99\"\r
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\n" ); document.write( "\n" ); document.write( "Now we know that the smallest f can be is 0 and the largest it could be is 100, because \"500%2A100=50000\". We can actually exclude 100 because that gets us to 50,000 with only 100 bills.\r
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\n" ); document.write( "\n" ); document.write( "The problem now becomes finding a whole number value for f so that the expression for h above comes out to be another whole number. You could make a list of all the results of \"%2849500-499f%29%2F99\" for each of the 100 possibilities for f, but that is a lot of calculator work. If you have access to MS Excel, you can solve it the way I did. I made a list of the numbers from 1 to 100, made a formula that calculated \"%2849500-499f%29%2F99\" for each of those numbers, and examined the list for an integer result.\r
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\n" ); document.write( "\n" ); document.write( "There is an analysis method, but it is a bit convoluted. First thing to notice is that \"49500%2F99=500\", an integer, therefore we only need to find f so that \"499f%2F99\" is an integer. Well, right from the start we can see that \"f=0\" is a possibility, but I think we can exclude that result because the wording of the problem at least implies that he has at least one of each kind of bill.\r
\n" ); document.write( "\n" ); document.write( "Now notice that, in the case of \"f=1\", the remainder using integer division of \"499%2F99\" is 4. That means the remainder when \"f=2\" will be 8, and so on until you get to \"f=24\" where the remainder is 96. The next one, \"f=25\", the remainder reverts to 1 \"96%2B4-99=1\", and the 'by 4' count starts over. \"f=49\" has a remainder of 97, \"f=50\" => \"r=2\", \"f=74\" => \"r=98\", \"f=75\" => \"r=3\", \"f=98\" => \"r=95\", \"f=99\" => \"r=0\"!!!! And we have found the answer.\r
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\n" ); document.write( "\n" ); document.write( "Turns out there is only one integer result, excluding the trivial result of h = 500, f = 0, and s = 0, and that is: h = 1, f = 99, and s = 400.
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