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\n" ); document.write( "The response from the other tutor shows a solution using formal algebra for the whole problem.
\n" ); document.write( "The algebraic part of the solution is much easier if you use some logical reasoning at the beginning.
\n" ); document.write( "The total value of the $5 and $10 notes is a multiple of $5; the total value of the 13 notes is $67, which is $2 more than a multiple of $5. So the total value of the $2 notes can only be $2, or $12, or $22, or.... Since the number of notes is 13, the only possibilities are for the total value of the $2 notes to be either $2 or $12.
\n" ); document.write( "So treat the problem as two distinct easier problems, using either one or six $2 notes.
\n" ); document.write( "(1) Using one $2 note....
\n" ); document.write( "The remaining amount, using 12 $5 and $10 notes, is $67-$2=$65. With 12 $5 and $10 notes, the total can be any multiple of $5 between 12($5)=$60 and 12($10)=$120. Since $65 is between $60 and $120, there is a solution using one $2 note.
\n" ); document.write( "Use mental arithmetic or formal algebra to find that the solution with one $2 note uses eleven $5 notes and one $10 note.
\n" ); document.write( "ANSWER #1: one $2 note, eleven $5 notes, and one $10 note. 1+11+1 = 13; $2+$55+10=$67.
\n" ); document.write( "(2) Using six $2 notes....
\n" ); document.write( "The remaining amount, using 7 $5 and $10 notes, is $67-$12=$55. With 7 $5 and $10 notes, the total can be any multiple of $5 between 7($5)=$35 and 7($10)=$70. Since $55 is between $35 and $70, there is also a solution using six $2 notes.
\n" ); document.write( "Again use mental arithmetic or formal algebra to find that the solution with six $2 notes uses three $5 notes and four $10 notes.
\n" ); document.write( "ANSWER #2: six $2 notes, three $5 notes, and four $10 notes. 6+3+4=13; $12+$15+$40=$67.
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\n" ); document.write( "Added after seeing responses from other tutors....
\n" ); document.write( "Responses from other tutors show some form of the equation 8x+3y=41 and then show different ways of finding non-negative integer solutions. Below I show a different way of finding all the solutions, starting from that equation.
\n" ); document.write( "In the equation 8x+3y=41, x and y are integers. 8x is even, and 41 is odd, so 3y must be odd; that means y is odd.
\n" ); document.write( "To find the \"first\" solution, use trial and error, using the smallest odd numbers for y, to find the smallest odd y that produces an integer value for x.
\n" ); document.write( "y=1: 8x+3=41; 8x=38 --> No...
\n" ); document.write( "y=3: 8x+9=41; 8x=32 --> Yes, x=4
\n" ); document.write( "So one solution is with y=3 and x=4; then, since x+y+z=13, z=6.
\n" ); document.write( "ANSWER #1: (x,y,z) = (4,3,6)
\n" ); document.write( "Now to find the other solutions....
\n" ); document.write( "The coefficients 8 and 3 in the equation 8x+3y=41 are relatively prime. When that is the case, all other solutions can be found by either adding 3 to x and subtracting 8 from y, or by subtracting 3 from x and adding 8 to y.
\n" ); document.write( "Our first solution was with y=3, so we can't subtract 8 from y to get another solution. But we can add 8 to y=3 to get y=11; when we do that, we subtract 3 from x=4 to get x=1. That gives us a second solution. With x=1 and y=11, x+y+z=13 gives us z=1.
\n" ); document.write( "ANSWER #2: (x,y,z) = (1,11,1)
\n" ); document.write( "We can't find other answers in non-negative integers by that method, so the two solutions we have found are the only ones.
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