A coin collection consists of nickels, dimes and quarters. There are 20 coins in the collection with a total value $2.00, and the number of nickels is three times more than the number of quarters.
Find all possible solutions of the numbers of coins of each kind.
[Problem submitted by Vick Vardanyan, LACC Instructor of Mathematics.]
Solution for Problem 1:
Let n be the number of nickels, d be the number of dimes, q the number of quarters.
Then n+d+q=20 (1)
Substituting (3) into (1) and (2) and simplifying one gets the same equation: d+4q=20.
But the possible positive integer solutions are
q=1, d=16, n=3
q=2, d=12, n=6
q=3, d=8, n=9
q=4, d=4, n=12.