Final answer:
The least positive integer m required to make the product m · 2! · 3! · ... · 16! a perfect square is 7. This is because the least frequent odd prime in the prime factorization of this product is 7, and multiplying by another 7 ensures all primes have an even exponent, resulting in a perfect square.
Step-by-step explanation:
The student is asking for the least positive integer m such that m multiplied by the product of factorials from 2! up to 16! forms a perfect square. A perfect square is an integer that is the square of an integer. To solve this, we must make sure that each prime factor in the product has an even exponent, which is required for a number to be a perfect square.
Steps to Solve:
Notice that in the factorial chain from 2! to 16!, each factorial includes the products of all smaller factorials. Thus, 16! includes all the multiples from 2 to 16.
- Recognize that any prime factorization within this range will occur many times, and its exponent in the overall product will be the sum of these frequencies.
- Calculate or recall the exponents of each prime factor. To achieve a perfect square, each in the prime factorization must have an even exponent.
- For primes greater than 8, their exponents are already even since they appear at most once in each factorial and will only contribute once their square is reached (i.e., 11 only in 11!, 13 only in 13!).
- Primes 2, 3, 5, and 7 require attention. Since 2 is the most abundant prime in factorials, identify the least frequent among the odd primes (3, 5, 7) and find the minimum integer m that ensures an even exponent for that prime.
- Upon examination, 7 requires the highest adjustment as it's the least frequent in the sequence, appearing with an even exponent first in 14! (two 7s) and with an odd exponent in 7!, thus needing another 7.
- Therefore, the least positive integer m that can be multiplied to ensure a perfect square is just another factor of 7, making m=7.