To find the largest number of groups Miki can create where each group has the same number of nickels and the same number of dimes, we need to find the greatest common divisor (GCD) of the number of nickels (104) and the number of dimes (88).
Let's find the GCD of 104 and 88.
First, we can list the prime factorizations of each number:
-
104 can be factored as: \[ 104 = 2 \times 52 = 2 \times 2 \times 26 = 2 \times 2 \times 2 \times 13 = 2^3 \times 13 \]
-
88 can be factored as: \[ 88 = 2 \times 44 = 2 \times 2 \times 22 = 2 \times 2 \times 2 \times 11 = 2^3 \times 11 \]
Now, we take the smallest power for each prime factor that appears in both factorizations:
- For \(2\), the minimum power is \(2^3\) (both have \(2^3\)).
- \(13\) appears only in the factorization of \(104\).
- \(11\) appears only in the factorization of \(88\).
Thus, the GCD is \(2^3 = 8\).
This means the largest number of groups Miki can have, where each group has the same number of nickels and dimes, is 8. Therefore, the answer is:
\[ \boxed{8} \]