How many factors of 24 × 35 × 104 are perfect squares which are greater than 1?

Concept: A positive integer's prime factorization determines its perfect-square divisors: a divisor is a perfect square exactly when every prime in it appears with an even exponent (0, 2, 4, ... up to that prime's exponent in the number). The count of perfect-square divisors is the product, over each prime, of (half that prime's exponent, rounded down, plus 1); subtract 1 to exclude the trivial divisor 1 when only values greater than 1 are wanted.

Applying to this question:

1. Rewrite 104 as (2 × 5)4 = 24 × 54, so the number is 24 × 35 × 104 = 24 × 35 × 24 × 54 = 28 × 35 × 54.

2. For a divisor 2a × 3b × 5c to be a perfect square, each of a, b, and c must individually be even.

3. Since the exponent of 2 is 8, a can be any even value from 0 up to 8 (i.e. 0, 2, 4, 6, 8).

4. Since the exponent of 3 is 5, b can be any even value from 0 up to 5 (i.e. 0, 2, 4).

5. Since the exponent of 5 is 4, c can be any even value from 0 up to 4 (i.e. 0, 2, 4).

6. Multiplying the number of choices for a, b, and c together gives the total number of perfect-square divisors, including the divisor equal to 1.

7. Excluding the trivial divisor 1 (the case a = b = c = 0) leaves the perfect-square factors that are greater than 1.

Cross-check: The direct formula for counting perfect-square divisors — multiplying (⌊exponent/2⌋ + 1) for every prime in the factorization — gives the same product of choices computed above, confirming the step-by-step count; removing the trivial divisor 1 again gives the same result.

Answer: There are 44 perfect-square factors of the number that are greater than 1.