M303 20260425 Number Theory (Part 2)
2003 AMC 12A Problems/Problem 23
How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$?
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$
#Number_of_Divisors_Function
Definition
$d(n)$ represents the number of divisors of $n$.
- Example: $6$ has 4 divisors ($1, 2, 3, 6$).
- Therefore, $d(6) = 4$.
The Formula
Suppose the prime factorization of $n$ is:
$$n = p_1^{e_1} p_2^{e_2} \dots p_i^{e_i}$$
Then the number of divisors is:
$$d(n) = (e_1 + 1)(e_2 + 1) \dots (e_i + 1)$$
Proof Summary
The proof uses combinatorics (counting choices):
1. Any divisor $d$ must be composed of the same prime factors as $n$:
$$d = p_1^{x_1} p_2^{x_2} \dots p_i^{x_i}$$
2. For each prime $p$, the exponent $x$ can be any integer from $0$ to the maximum exponent $e$ found in $n$.
- For $p_1$, there are $(e_1 + 1)$ choices ($0, 1, 2, \dots, e_1$).
- For $p_2$, there are $(e_2 + 1)$ choices.
- ...and so on.
3. By the multiplication principle, the total number of ways to form a divisor is the product of these choices.
Counting Perfect Square Divisors
A number is a perfect square if and only if every exponent in its prime factorization is even.
For a divisor $d$ to be a perfect square, it must take the form:
$$d = p_1^{2x_1} p_2^{2x_2} \dots p_i^{2x_i}$$
This means we are restricted to choosing only even exponents ($0, 2, 4, \dots$) for each prime factor.
Number of Choices
For each prime factor $p_i$ with an original exponent $e_i$ in the number $n$:
1. We need to find how many even numbers exist between $0$ and $e_i$.
2. The formula for this count is:
$$\lfloor \frac{e_i}{2} \rfloor + 1 \text{ choices}$$
(Note: The symbol $\lfloor \rfloor$ denotes the floor function, which means rounding down to the nearest whole number).
Summary
The total number of divisors of $n$ that are also perfect squares is the product of these choices:
$$\text{Total Square Divisors} = (\lfloor \frac{e_1}{2} \rfloor + 1)(\lfloor \frac{e_2}{2} \rfloor + 1) \dots (\lfloor \frac{e_i}{2} \rfloor + 1)$$
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Example for Clarity
If $n = 72$, its prime factorization is $2^3 \cdot 3^2$.
- For $2^3$, the exponent is $e=3$. Even choices are $0$ and $2$.
- Formula: $\lfloor 3/2 \rfloor + 1 = 1 + 1 = \mathbf{2}$ choices.
- For $3^2$, the exponent is $e=2$. Even choices are $0$ and $2$.
- Formula: $\lfloor 2/2 \rfloor + 1 = 1 + 1 = \mathbf{2}$ choices.
- Total perfect square divisors: $2 \times 2 = 4$.
(They are $1$, $4$, $9$, and $36$).
2005 AMC 12B Problems/Problem 21
A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$?
$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$