.
You have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In his post, tutor @Theo found out one solution " 10 pennies, 12 nickels, 3 dimes and 0 quarters".
He used the trial and error guessing.
But the problem has two other solutions that eluded Theo.
Below I give a complete analysis to catch all possible solutions.
Let x be the number of pennies,
y be the number of nickels,
z be the number of dimes,
w be the number of quarters.
We have two equations
x + y + z + w = 25, (1) for the total number of coins
x + 5y + 10z + 25w = 100 (2) for the total value.
From equation (2), subtract equation (1). You will get
4y + 9z + 24w = 75,
4y + 24w = 75 - 9z. (3)
In the last equation, left side value is always a multiple of 4.
Hence, right side must be a multiple of 4.
z = 0 gives for the right side of (3) the value 75 - 9*0 = 75, which is not a multiple of 4.
z = 1 gives for the right side of (3) the value 75 - 9*1 = 66, which is not a multiple of 4.
z = 2 gives for the right side of (3) the value 75 - 9*2 = 57, which is not a multiple of 4.
z = 3 gives for the right side of (3) the value 75 - 9*3 = 48, which is a multiple of 4.
At z = 3, from (3), we have
4y + 24w = 48. (3)
One possible solution for (3) is (y,w) = (0,2). Then x + y + z + w = x + 0 + 3 + 2 = 25 implies x = 20.
So, one possible solution for the problem is 20 pennies, 0 nickels, 3 dimes and 2 quarters.
Second possible solution for (3) is (y,w) = (6,1). Then x + y + z + w = x + 6 + 3 + 1 = 25 implies x = 15.
So, second possible solution for the problem is 15 pennies, 6 nickels, 3 dimes and 1 quarters.
Third possible solution for (3) is (y,w) = (12,0). Then x + y + z + w = x + 12 + 3 + 0 = 25 implies x = 10.
So, third possible solution for the problem is 10 pennies, 12 nickels, 3 dimes and 0 quarters.
Thus the problem has three possible solutions:
(1) 20 pennies, 0 nickels, 3 dimes and 2 quarters,
(2) 15 pennies, 6 nickels, 3 dimes and 1 quarters,
(3) 10 pennies, 12 nickels, 3 dimes and 0 quarters.
It is clear that there no other solutions.
Solved completely.
-----------------------------
Interesting fact:
Today, November 10, 2025, after completing my solution, I submitted this problem to two artificial intelligence sites.
I was guided by my curiosity: it was interesting to me, how they will treat this problem ?
One AI was Google AI Overview. The second AI was math-gpt.org
Both produced incomplete answers.
Google AI Overview referred to the incomplete @Theo solution.
Math-gpt.org did not share his links with me.
But I made my conclusion: they both can not think (in the usual meaning of this word)
- they both re-write (compile) from existing solutions in their databases.
Surely, as usual, I informed the Google AI Overview about their mistake through their
feedback system with the reference to the current link.
.
You have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let x be the number of pennies,
y be the number of nickels,
z be the number of dimes,
w be the number of quarters.
We have two equations
x + y + z + w = 25, (1) for the total number of coins
x + 5y + 10z + 25w = 100 (2) for the total value.
From equation (2), subtract equation (1). You will get
4y + 9z + 24w = 75,
4y + 24w = 75 - 9z. (3)
In the last equation, left side value is always a multiple of 4.
Hence, right side must be a multiple of 4.
z = 0 gives for the right side of (3) the value 75 - 9*0 = 75, which is not a multiple of 4.
z = 1 gives for the right side of (3) the value 75 - 9*1 = 66, which is not a multiple of 4.
z = 2 gives for the right side of (3) the value 75 - 9*2 = 57, which is not a multiple of 4.
z = 3 gives for the right side of (3) the value 75 - 9*3 = 48, which is a multiple of 4.
At z = 3, from (3), we have
4y + 24w = 48. (3)
One possible solution for (3) is (y,w) = (0,2). Then x + y + z + w = x + 0 + 3 + 2 = 25 implies x = 20.
So, one possible solution for the problem is 20 pennies, 0 nickels, 3 dimes and 2 quarters.
Second possible solution for (3) is (y,w) = (6,1). Then x + y + z + w = x + 6 + 3 + 1 = 25 implies x = 15.
So, second possible solution for the problem is 15 pennies, 6 nickels, 3 dimes and 1 quarters.
Third possible solution for (3) is (y,w) = (12,0). Then x + y + z + w = x + 12 + 3 + 0 = 25 implies x = 10.
So, third possible solution for the problem is 10 pennies, 12 nickels, 3 dimes and 0 quarters.
Thus the problem has three possible solutions:
(1) 20 pennies, 0 nickels, 3 dimes and 2 quarters,
(2) 15 pennies, 6 nickels, 3 dimes and 1 quarters,
(3) 10 pennies, 12 nickels, 3 dimes and 0 quarters.
It is clear that there no other solutions.
Solved completely.