To find the largest number of groups Mikey can form with the same number of nickels and the same number of dimes in each group, we need to determine the greatest common divisor (GCD) of the number of nickels (104) and the number of dimes (88).

We can use the Euclidean algorithm to find the GCD.

1. First Step: Divide 104 by 88 and find the remainder. $104 \div 88 = 1 \quad \text{remainder} \quad 104 - 88 \cdot 1 = 104 - 88 = 16$ So, the remainder is 16.

2. Second Step: Now, we apply the algorithm with 88 and 16. $88 \div 16 = 5 \quad \text{remainder} \quad 88 - 16 \cdot 5 = 88 - 80 = 8$ So, the remainder is 8.

3. Third Step: Now, we apply the algorithm with 16 and 8. $16 \div 8 = 2 \quad \text{remainder} \quad 16 - 8 \cdot 2 = 16 - 16 = 0$ So, the remainder is 0.

When we reach a remainder of 0, the divisor at this step (which is 8) is the GCD of 104 and 88.

Therefore, the largest number of groups Mikey can have, where each group has the same number of nickels and the same number of dimes, is:

$\boxed{8}$

Would you like more details or have any questions?

Here are 5 related questions to expand on this information:

1. How many nickels and dimes would be in each group if Mikey divides her coins into the maximum number of groups?
2. What is the process to find the greatest common divisor (GCD) of two numbers using the Euclidean algorithm?
3. If Mikey had 120 nickels and 90 dimes, what would be the largest number of groups she could form?
4. How does the least common multiple (LCM) of two numbers relate to the greatest common divisor (GCD)?
5. Can you find the GCD of 104 and 88 using prime factorization?

Tip: To verify your answer when using the Euclidean algorithm, you can backtrack by multiplying the divisors and checking if the sequence correctly reconstructs the original numbers.