Question 1185540
<br>
A pure mathematical solution to this problem would be very complex; however, there are many ways to solve this easily using logical reasoning.<br>
You will learn nothing from this, and get no valuable mental exercise, if we simply show you the answer, or even if we show you the complete solution process.<br>
I will instead suggest a few ways you can start on the problem and let you finish.  To get the most benefit, try finishing the solution from each of the starts I show.<br>
(1) A "greedy" approach -- starting with the highest value coins....<br>
We can't have two ₱10 coins, because that would make ₱20, and we would have only ₱0.50 to make with the remaining 7 coins.  So we can only have one ₱10 coin.
That leaves ₱10.50 to make with the other 8 coins.
A similar analysis then shows we can only have one ₱5 coin.<br>
...then continue from there....<br>
(2) Using the fact that only the ₱0.25 coins have values that are fractions....<br>
With a total of ₱20.50, the number of ₱0.25 coins must be either 2 (to make ₱0.50) or 6 (to make ₱1.50). Quick analysis shows there can't be 6 ₱0.25 coins, because that would leave only 3 coins to make the remaining ₱19.  So the number of ₱0.25 coins has to be 2.<br>
... and again continue from there....<br>
(3) Using the assumption that the statement of the problem implies that he has at least one of each coin....<br>
1 of each coin make a total value of ₱16.25, leaving 5 more coins to make the remaining ₱4.25.  Obviously one of the remaining coins has to be another ₱0.25 coin....<br>
... and then continue on that path....<br>