Question 1207307
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Alan has 13 pieces of $2 $5 and $10 notes in his wallet. 
If the amount of money he has is $67 how many pieces of each note does he have ?
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        In this my post,  I'd like to present a regular solution to this problem.

        Not the most witty and not the most slow,  but something the most  REGULAR, 

        which is in between of these extremes.



<pre>
Let the numbers of pieces of $2, $5, and $10 notes be, respectively, x, y, and z. 
Then you have this system of two equations in three unknowns

     x  + y +  z  = 13,     (1)
    2x + 5y + 10z = 67.     (2)


You should find a solution / (solutions) in integer non-negative numbers.


Multiply first equation by 2

    2x + 2y +  2z = 26,     (1')
    2x + 5y + 10z = 67.     (2)


Subtract equation (1') from equation (2).  You will get

         3y +  8z = 41.     (3)


This equation is in two unknowns, but you need to solve it in integer non-negative numbers.


Under this restriction, this equation is so called linear Diophantine equation.
A standard way to solve it is "trial and error".
In other words, you should try several integer positive values of z, 
and find relevant values of y. Those values of z that provide non-negative integer y will be the solutions.

From equation (3), the candidates for z to try are z = 1, 2, 3, 4, and 5 - - - only five values,
so it is not a catastrophic job to try all five.


To facilitate this job, people usually make a Table such as I provide below


    z        41-8z              is y = 41-8z           Integer solution
                                a multiple of 3 ?      y = {{{(41-8z)/3}}}
    ----------------------------------------------------------------------------

    1   41-8*1 = 41- 8 = 33           Yes                  11

    2   41-8*2 = 41-16 = 25           No

    3   41-8*3 = 41-24 = 17           No

    4   41-8*4 = 41-32 =  9           Yes                   3

    5   41-8*5 = 41-40 =  1           No


From the Table, you see that there are exactly two solutions to equation (3) for y and z.

One solution is        (y,z) = (11,1).  The corresponding value of x, from (1), is x= 13 - y - z = 13 - 11 - 1 = 1.

The other solution is  (y,z) =  (3,4).  The corresponding value of x, from (1), is x= 13 - y - z = 13 - 3 - 4 = 6.


At this point, the problem is just solved, in full.


<U>ANSWER</U>.  There are two solutions.
         One solution is       (one $2 note,  eleven $5 notes and one  $10 note).
         The other solution is (six $2 notes, three  $5 notes and four $10 notes).


You can easily CHECK it on your own that these numbers satisfy to all problem's requirements.
</pre>

Solved.


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In this problem, making "trial and errors" was dowable procedure.


In other similar problems, if the number of "trials and errors" is great, you may use
the tools like Excel to facilitate such a job and to make a calculation Table quickly.


In any case, what is described in my post, is traditionally considered as a standard procedure
to solve similar problems.


Again, as a reminder, originally you have only two equations for three unknowns.
But an additional restriction of having non-negative integer solutions reduces 
the possible number of solutions from infinity to a finite number.