There are 52 coins that consists of pennies, dimes, and quarters. Two stacks have the same number of coins and the third stack has twice as much. The coins equal to $5.98. How many pennies, dimes, and quarters are there? I already found out that two stacks have 13 coins and one stack has 26 coins.

Since you already have the second part, let's work on the number of pennies , dimes and quarters we have to make $5.98

number of pennies -- x
number of dimes ---- y
number of quarters -- 52-x-y

x + 10y + 25(52-x-y) = 598
x+10y + 1300 - 25x - 25y = 598
-24x - 15y = -702
8x + 5y = 234

y = (234 - 8x)/5

by trial and error, I found the smallest integer value to work is
x = 3, y = 34
since the slope of 8x+5y=234 is -8/5 ,
increasing the x by 8 and decreasing the y by 5 has to give me another solution,
So I got the following:

P D Q

3 34 10
13 26 13
18 18 16
23 10 19
28 2 20

So there are 5 different solutions to your problem
testing each, you can see that you get a sum of $5.98 for each one.

e.g. 23 pennies, 10 dimes and 19 quarters
= 23 + 10(10) + 25(19) = 598

Your question does not specify what type of coins each stack must contain.

An interesting addition to the problem would be to insist that each stack consists of the same denomination of coins in the ratio of 1 : 1 : 2
Would this be even possible??

To find the number of pennies, dimes, and quarters, we'll use the information given.

Let's assume the number of pennies in each stack is p, the number of dimes is d, and the number of quarters is q.

We know the following:

- The two stacks with 13 coins each have a total of 2 * 13 = 26 coins.
- The stack with 26 coins has twice the number of coins, so its total is 2 * 26 = 52 coins.

Based on the above information, we can write three equations and solve them simultaneously.

1. Number of coins equation:
p + d + q = 52

2. Equation for the first two stacks:
13p + 13d + 13q = 26

3. Equation for the third stack:
26p + 26d + 26q = 52

Now, let's solve these equations.

From equation 2, we can divide both sides by 13:
p + d + q = 2

Substitute this value of (p + d + q) in equation 1:
2 = 52
p + d + q = 2

Now, we can add equations 1 and 2:
2(p + d + q) = 52 + 2
2p + 2d + 2q = 54

Divide both sides by 2:
p + d + q = 27

This means that:
p = 27 - d - q

Substitute this value of p in equation 3:
26(27 - d - q) + 26d + 26q = 52

Simplify the equation:
702 - 26d - 26q + 26d + 26q = 52

The -26d and +26d terms cancel each other out, as do the -26q and +26q terms:
702 = 52

This equation is not possible, as 702 does not equal 52. Therefore, there is no solution that satisfies the given conditions.

Please double-check the provided information or provide additional details to assist you further in finding a solution.

To solve this problem, we can represent the number of pennies, dimes, and quarters in each stack using variables. Let's call the number of pennies 'p', the number of dimes 'd', and the number of quarters 'q'.

From the information given, we know that there are two stacks with 13 coins each, and one stack with 26 coins. Therefore, we can set up the following equations:

Stack 1: p + d + q = 13
Stack 2: p + d + q = 13
Stack 3: p + d + q = 26

Next, we need to consider the value of the coins. Since the value of a penny is $0.01, the value of a dime is $0.10, and the value of a quarter is $0.25, we can use these values to set up another equation based on the total value of the coins:

0.01p + 0.1d + 0.25q = 5.98

Now we have a system of four equations:

p + d + q = 13
p + d + q = 13
p + d + q = 26
0.01p + 0.1d + 0.25q = 5.98

We can solve this system of equations using substitution or elimination. Let's use elimination.

Subtracting the first equation from the second equation, we get:

(p + d + q) - (p + d + q) = 13 - 13
0 = 0

This means that the first two equations are equivalent, so we can ignore the second equation. Now we have a system of three equations:

p + d + q = 13
p + d + q = 26
0.01p + 0.1d + 0.25q = 5.98

Subtracting the first equation from the third equation, we get:

(0.01p + 0.1d + 0.25q) - (p + d + q) = 5.98 - 13
-0.99p - 0.9d - 0.75q = -7.02

Now we have a system of two equations:

p + d + q = 13
-0.99p - 0.9d - 0.75q = -7.02

We can solve this system of equations using substitution or elimination methods to find the values of p, d, and q, which represent the number of pennies, dimes, and quarters, respectively.

so the answer is 28 pennies 7 dimes and 5 sets of quarters