Hello, jmc1979!
Is it possible to have 50 coins, all of which are pennies, dimes, or quarters, with a total worth $3?
Let P = number of pennies.
Let D = number of dimes.
Let Q = number of quarters.
A total of 50 coins: . [1] . . P + D + Q . . = . 50
Total value of 300¢: .[2] .P + 10D + 25Q .= .300
From [1], we have: .Q .= .50 - P - D
Substitute into [2]: .P + 10D + 25(50 - P - D) .= .300
This simplifies to: .24P + 15D .= .950 . → . 3(8P + 5D) .= .950
The left side is a multiple of 3 . . . The right side is not.
. . An impossible situation . . .
The answer to the question is: .No
Curses . . . too slow again!