1. ## Linear Diophantine Equations

I am having difficulties with this problem:

Is it possible to have 50 coins, all of which are pennies, dimes, or quarters, with a total worth $3? Thank you! 2. Originally Posted by jmc1979 I am having difficulties with this problem: Is it possible to have 50 coins, all of which are pennies, dimes, or quarters, with a total worth$3?

Thank you!
Let,
x # of pennies.
y # of dimes.
z # of quaters.

Henceforth,
x+y+z=50
x+10y+25z=300

Thus,
(50-y-z)+10y+25z=300
9y+24z=250
Note that,
gcd(9,24)=3 and 3 does not divide 250.
Thus, there can be no such combination of coins.

3. 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!