how about 18 nickels and 1 dime
or
9 dimes and 10 penny
or
..
suppose we restrict ourselves to p pennies, n nickels, d dimes, q quarter
then we need integer solutions to
pennies + nickels = 100
p + 5(19-p) = 100
p + 95 -5p = 100
-4p = 5 --->no integer solution
pennies + dimes = 100
p + 10(19-p) = 100
p + 190 - 10p = 100
-9p = -90
p = 10 ---> 10 pennies, and 19-10 or 9 dimes, see above
pennies + quarters = 100 ?
25q + 1(19-q) = 100
-24q = 81 , no way
nickels and dimes = 100 ?
5n + 10(19-n) = 100
5n + 190 - 10n = 100
-5n = -90
n = 18 ---> 18 nickels, and 1 dime , see above
nickels + quarters ?
5n + 25(19-n) = 100
-20n = 375 , no integer for n
dimes and quarters ??
10d + 25(19-d) = 100
-15d = -375
d = 25 , but we only had 19 coins, so no way
looks like we only have the two solutions stated above
Jeremy has 19 coins of 2 different values. the total value of the coins is $1.00. What are the coins? How many of each coin does he have?
1 answer