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We know that, any factor of the (25 × 33 × 52) can be expressed as 2a × 3b × 5c, where a ranges from 0 to 5, b ranges from 0 to 3 and c ranges from 0 to 2. Here, we need to find even factors of 25× 33× 52, so there has to be at least one factor of 2 in the given number. So, a will range from 1 to 5. As a result, a can take 5 values, b can take 4 values and c can take 3 values. ∴ Total number of even factors = 5 × 4 × 3 = 60
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