Question 1207307
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Hi
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

Let number of $2 and $5 notes be T, and F, respectively
Since there are 13 notes, number of $10 notes = 13 - T - F
We then get the total-amount equation as: 2T + 5F + 10(13 - T - F) = 67
                                         2T + 5F + 130 - 10T - 10F = 67
                                                         - 8T - 5F = 67 - 130  
                                                           8T + 5F = 63
Now, as 3 UNKNOWNS/VARIABLES are being sought, but an equation with 2 variables is derived, we use the
DIOPHANTINE-EQUATIION method to solve. 
8T + 5F = 63 ----- eq (i)
Because MULTIPLES of 5 have units digits of either 0 or 5, we solve eq (i) in terms of F, so our final
equation will have a denominator of 5
8T + 5F = 63 then becomes: {{{matrix(2,3, 5F, "=", 63 - 8T, F, "=", (63 - 8T)/5)}}}
Now, looking at the above equation, we see that, substituting values of T, from 1, up to 7 - NOT 8,
because 63 - 8(8) will produce a NEGATIVE number of $5 bills (UNACCEPTABLE!!) in the numerator. Now, 
again looking at {{{matrix(1,3, F, "=", (63 - 8T)/5)}}}, we see that the numerator 63 - 8T MUST produce a number that is a 
MILTIPLE of 5, but 5's MULTIPLES ONLY end in 0 or 5. Therefore, the numbers from 1 - 7 that MUST be 
multiplied by the 8 in 8T, MUST PRODUCE a number that, when SUBTRACTED from 63, gives a MULTIPLE of
5 (the denominator) with units digit of either 0 or 5. For this numerator-MULTPLE of 5 to end
in 0, a MULTIPLE of 8 would need to end with a units digit of 3 (63 - ?3 = ?0). But, there is NO
MULTIPLE of 8 that end in 3. AS a matter of fact, 8 is an even number and when multiplied by any 
other number (odd or even), another EVEN number will ensue. So, the only other multiples of 8 would
be those that end in 8 (have a units digit of 8). When checked, digits from 1 - 7, produce only 2
multiples of 8 that end with an 8. These are 1 (8 * 1 = 8), and 6 (8 * 6 = 48). Let's now apply these 
values of T to the DIOPHANTINE equation, {{{matrix(1,3, F, "=", (63 - 8T)/5))}}}
    With T = 1, we get: {{{matrix(2,3, F, "=", (63 - 8(1))/5, F, "=", (63 - 8)/5)}}}                       With T = 6, we get: {{{matrix(2,3, F, "=", (63 - 8(6))/5, F, "=", (63 - 48)/5)}}}   
Number of $5 notes, or {{{highlight_green(matrix(1,5, F, "=", 55/5, "=", 11))}}}                     Number of $5 notes, or {{{highlight_green(matrix(1,5, F, "=", 15/5, "=", 3))}}}
As such, number of $10 notes = 13 - 1 - 11 = 1            As such, number of $10 notes = 13 - 6 - 3 = 4  
This gives us: {{{highlight_green(system(matrix(3,5, Number, of, highlight("$2"), "bills:", highlight(1),
Number, of, highlight("$5"), "bills:", highlight(11),
Number, of, highlight("$10"), "bills:", highlight(1))))}}}           This gives us: {{{highlight_green(system(matrix(3,5, Number, of, highlight("$2"), "bills:", highlight(6),
Number, of, highlight("$5"), "bills:", highlight(3),
Number, of, highlight("$10"), "bills:", highlight(4))))}}}</pre>