Question 1207307
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The response from the other tutor shows a solution using formal algebra for the whole problem.<br>
The algebraic part of the solution is much easier if you use some logical reasoning at the beginning.<br>
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.<br>
So treat the problem as two distinct easier problems, using either one or six $2 notes.<br>
(1) Using one $2 note....<br>
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.<br>
Use mental arithmetic or formal algebra to find that the solution with one $2 note uses eleven $5 notes and one $10 note.<br>
ANSWER #1: one $2 note, eleven $5 notes, and one $10 note. 1+11+1 = 13; $2+$55+10=$67.<br>
(2) Using six $2 notes....<br>
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.<br>
Again use mental arithmetic or formal algebra to find that the solution with six $2 notes uses three $5 notes and four $10 notes.<br>
ANSWER #2: six $2 notes, three $5 notes, and four $10 notes. 6+3+4=13; $12+$15+$40=$67.<br>
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Added after seeing responses from other tutors....<br>
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.<br>
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.<br>
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.<br>
y=1: 8x+3=41; 8x=38 --> No...
y=3: 8x+9=41; 8x=32 --> Yes, x=4<br>
So one solution is with y=3 and x=4; then, since x+y+z=13, z=6.<br>
ANSWER #1: (x,y,z) = (4,3,6)<br>
Now to find the other solutions....<br>
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.<br>
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.<br>
ANSWER #2: (x,y,z) = (1,11,1)<br>
We can't find other answers in non-negative integers by that method, so the two solutions we have found are the only ones.<br>