The product \(64 \times 63 \times 62 \times \cdots \times 3 \times 2 \times 1\) can be written as \(64!\) and called "\(64\) factorial".
In general, the product of the positive integers \(1\) to \(m\) is \[m!=m \times (m-1) \times (m-2) \times \cdots \times 3 \times 2 \times 1\]
If \(64!\) is divisible by \(2025^n\), determine the largest positive integer value of \(n\).
Let \(P=64!\). The prime factorization of \(2025\) is \(3^4 \times 5^2.\) We must determine how many times the factors of \(3\) and \(5\) are repeated in the factorization of \(P.\)
First we count the number of factors of \(3\) in \(P\) by looking at the multiples of \(3\) from \(1\) to \(64.\) They are \(3\), \(6\), \(9\), \(\ldots\), \(57\), \(60\), and \(63.\) Each of these \(21\) numbers contains a factor of \(3.\)
Now, each multiple of \(9\) from \(1\) to \(64\) will contain a second factor of \(3\). These multiples of \(9\) are \(9\), \(18\), \(27\), \(36\), \(45\), \(54\), and \(63.\) Each of these \(7\) numbers contains two factors of \(3.\)
Now, each multiple of \(27\) from \(1\) to \(64\) will contain a third factor of \(3\). These multiples of \(27\) are \(27\) and \(54.\) Each of these \(2\) numbers contains three factors of \(3.\)
There are no higher powers of \(3\) less than \(64.\) Thus, \(P\) has \(21+7+2 = 30\) factors of \(3\), and so the largest power of \(3\) that \(P\) is divisible by is \(3^{30}.\)
Next we count the number of factors of \(5\) in \(P\) by looking at the multiples of \(5\) from \(1\) to \(64.\) They are \(5\), \(10\), \(15\), \(\ldots\), \(50\), \(55\), and \(60.\) Each of these \(12\) numbers contains a factor of \(5.\)
Now, each multiple of \(25\) from \(1\) to \(64\) will contain a second factor of \(5.\) These multiples of \(25\) are \(25\) and \(50.\) Each of these \(2\) numbers contains two factors of \(5.\)
There are no higher powers of \(5\) less than \(64.\) Thus, \(P\) has \(12+2 = 14\) factors of \(5\), and so the largest power of \(5\) that \(P\) is divisible by is \(5^{14}.\)
Thus, \(P\) is divisible by \(3^{30} \times 5^{14}.\) \[3^{30} \times 5^{14} = 3^{28} \times 3^2 \times 5^{14} = \left(3^4 \times 5^2 \right)^7 \times 3^2 = 2025^7 \times 3^2\] Thus, \(P\) is divisible by \(2025^7\), and \(7\) is the largest value of \(n.\)