Natalie has some nickels, Dirk has some dimes, and Qunicy has some quarters. Dirk has five more dimes than Qunicy has quarters. If Natalie gives Dirk a nickel, Dirk gives Quincy a dimes, and Quincy gives Natalie a quarter, they will all have the same amount of money. How many coins did each have originally?

d = q+5

5(n-1)+25 = 10(d-1)+5 = 25(q-1)+10
(n,d,q) = (13,9,4)

To solve this problem, let's break it down step by step.

Let's assume that Natalie has "n" nickels, Dirk has "d" dimes, and Quincy has "q" quarters.

According to the given information, Dirk has five more dimes than Quincy has quarters, so we can set up an equation:

d = q + 5

Next, we are given the condition that if Natalie gives Dirk a nickel, Dirk gives Quincy a dime, and Quincy gives Natalie a quarter, they will all have the same amount of money.

If Natalie gives Dirk a nickel, Natalie will have n-1 nickels, and Dirk will have d+1 dimes.

If Dirk gives Quincy a dime, Dirk will have d-1 dimes, and Quincy will have q+1 quarters.

If Quincy gives Natalie a quarter, Quincy will have q-1 quarters, and Natalie will have n+1 nickels.

According to the given condition, they will all have the same amount of money. This means:

(n-1)(0.05) = (d+1)(0.10) = (q+1)(0.25)

Simplifying this equation, we get:

0.05n - 0.05 = 0.10d + 0.10 = 0.25q + 0.25

Combining the three parts of the equation, we get:

0.05n - 0.05 = 0.10d + 0.10 = 0.25q + 0.25

Now, let's solve the equation:

0.05n - 0.05 = 0.10d + 0.10 -> (1)

0.05n - 0.05 = 0.25q + 0.25 -> (2)

From equation (1), we have:

0.10d + 0.10 = 0.10n - 0.10

0.10d = 0.10n

d = n

From equation (2), we have:

0.25q + 0.25 = 0.10n - 0.10

0.25q = 0.10n - 0.35

5q = 2n - 7

From the equation d = n, we can substitute n for d in the equation:

5q = 2d - 7

Now, let's solve this equation to find the values of n, d, and q.

We can start by trying different values of q and checking if it satisfies the equation:

Let's assume q = 2.

Therefore,

5(2) = 2d - 7

10 = 2d - 7

2d = 17

d = 8.5

Since d cannot be a decimal, q cannot be 2.

Let's try the next possible value.

Let's assume q = 3.

Therefore,

5(3) = 2d - 7

15 = 2d - 7

2d = 22

d = 11

Now we know that q = 3 and d = 11.

Since d = n, we also know that n = 11.

Therefore, Natalie originally had 11 nickels, Dirk originally had 11 dimes, and Quincy originally had 3 quarters.