Question 1207307
<pre>

Let the numbers of pieces of $2, $5, and $10 notes be, respectively,
x, y, and z. 


{{{system(x+y+z=13,2x+5y+10z=67)}}}

Multiply the first equation by -2

{{{system(-2x-2y-2z=-26,2x+5y+10z=67)}}}

Add the equations term by term:

{{{3y+8z=41}}}

Write all the integers in terms of their nearest multiple of the
smallest absolute value of the coefficients of variables, 
which is 3:

{{{3y+(9-1)z=42-1}}}
{{{3y+9z-z=42-1}}}
Divide thru by 3

{{{y+3z-z/3=14-1/3}}}

Isolate the fraction terms 

{{{y+3z-14=z/3-1/3}}}

The left side is an integer, so the right side must also be an integer, 
say A

{{{system(y+3z-14=A,z/3-1/3=A)}}}

Multiply the second equation by 3

{{{system(y+3z-14=A,z-1=3A)}}}

{{{z=3A+1}}}

Substitute for z in y+3z-14=A:

{{{y+3(3A+1)-14=A}}}
{{{y+9A+3-14=A}}}
{{{y+9A-11=A}}}
{{{y=11-8A}}}

Substitute for y and z in

{{{x+y+z=13}}}

{{{x+11-8A+3A+1=13}}}
{{{x+12-5A=13}}}
{{{x=5A+1}}}

The general solution in integers is

{{{system(x=5A+1,y=11-8A,z=3A+1)}}}

x, y, and z must be non-negative integers:

{{{system(5A+1>=0,11-8A>=0,3A+1>=0)}}} => {{{system(A>=-1/5,A<=11/8,A>=-1/3)}}}

Thus from the above, A can only be 0 or 1

Choose A=0

x=1, y=11, z=1

Choose A=1

x=6, y=3, z=4

Two solutions. [Other choices for A will produce one or more negative answers.]

Edwin</pre>