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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 $n$.

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 $n$.

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:

$$\text{Total Square Divisors} = (\lfloor \frac{e_1}{2} \rfloor + 1)(\lfloor \frac{e_2}{2} \rfloor + 1) \dots (\lfloor \frac{e_i}{2} \rfloor + 1)$$

---

Example for Clarity

If $n = 72$, its prime factorization is $2^3 \cdot 3^2$.

(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 $n$?

$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$

#multiplicative_function

A function $f$ is considered multiplicative if, for any two numbers $m$ and $n$ that are relatively prime ($\gcd(m, n) = 1$), the following holds:

$$f(m \cdot n) = f(m) \cdot f(n)$$

2008 AMC 12B Problems/Problem 23

The sum of the base-$10$ logarithms of the divisors of $10^n$ is $792$. What is $n$?

$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$

2010 AMC 12B Problems/Problem 9

Let $n$ be the smallest positive integer such that $n$ is divisible by $20$, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

2011 AMC 12B Problems/Problem 15

How many positive two-digit integers are factors of $2^{24}-1$?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

#Legendres_Formula #p-adic_Valuation

Legendre's Formula p-adic valuation

This is a classic tool in number theory for finding how many times a prime number "fits" into a factorial.

1. Definition: $v_p(n)$

The function $v_p(n)$ is defined as the largest power of a prime $p$ that divides $n$.

2. The Formula for Factorials

To find the total exponent of a prime $p$ in the prime factorization of $n!$, we use:

$$v_p(n!) = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + \dots$$

(The sum continues until $p^k > n$, at which point the terms become 0).

3. Proof Sketch / Logic

The notes use $v_2(n!)$ as an example to explain why this works:

Visualizing the "Stairs":

The handwritten portion shows how multiples of 2, 4, 8, and 16 are layered. By summing these floor functions, you are effectively counting every instance of the prime $p$ across all numbers from $1$ to $n$.

Practical Example

If you wanted to find how many zeros are at the end of $100!$, you would need to find $v_5(100!)$:

1. $\lfloor 100/5 \rfloor = 20$

2. $\lfloor 100/25 \rfloor = 4$

3. $\lfloor 100/125 \rfloor = 0$

Total: $20 + 4 = 24$. So, $100!$ ends in 24 zeros.

This formula is extremely useful for AIME-level problems involving trailing zeros or divisibility of large factorials.

2015 AMC 10B Problems/Problem 23

Let $n$ be a positive integer greater than $4$ such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$. What is the sum of the digits of $s$?

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

2016 AMC 10A Problems/Problem 22

For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have?

$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$

#Greatest_Common_Divisor #Least_Common_Multiple

Key Relationship

For any two positive integers \(a\) and \(b\):

$\text{GCD}(a,b)\times \text{LCM}(a,b)=a\times b$

#Euclidean_Algorithm

The Euclidean algorithm is an efficient method for finding the Greatest Common Divisor (GCD) of two integers by repeatedly dividing the larger number by the smaller one and replacing the larger number with the remainder.

Euclidean algorithm

The basic idea is to repeatedly use the fact that $\gcd({a,b}) \equiv \gcd({b,a - b})$

Example

To see how it works, just take an example. Say $a = 93, b=42$.

We have $93 \equiv 9 \pmod{42}$, so $\gcd(93,42) = \gcd(42,9)$.

Similarly, $42 \equiv 6 \pmod{9}$, so $\gcd(42,9) = \gcd(9,6)$.

Continuing, $9 \equiv 3 \pmod{6}$, so $\gcd(9,6) = \gcd(6,3)$.

Then $6 \equiv 0 \pmod{3}$, so $\gcd(6,3) = \gcd(3,0) = 3$.

Thus $\gcd(93,42) = 3$.

$93 = 2 \cdot 42 + 9 \qquad (1)$

$42 = 4 \cdot 9 + 6 \qquad (2)$

$9 = 1 \cdot 6 + 3 \qquad (3)$

$6 = 2 \cdot 3 + 0 \qquad (4)$

Extended Euclidean Algorithm

An added bonus of the Euclidean algorithm is the "linear representation" of the greatest common divisor. This allows us to write $\gcd(a,b)=ax+by$, where $x,y$ are some elements from the same Euclidean Domain as $a$ and $b$ that can be determined using the algorithm. We can work backwards from whichever step is the most convenient.

Continuing the previous example, our goal is to find $a$ and $b$ such that $93a + 42b = \gcd(93,42) = 3.$ We can work backwards from equation $(3)$ since $3$ appears there:

$3 = 9 - 1 \cdot 6$

We currently have $3$ as a linear combination of $6$ and $9$. Our goal is to replace $6$ and $9$ so that we have a linear combination of $42$ and $93$ only. We start by rearranging $(2)$ to $6 = 42 - 4 \cdot 9$ so we can substitute $6$ to express $3$ as a linear combination of $9$ and $42$:

$3 = 9 - 1 \cdot (42 - 4 \cdot 9)$

$3 = -1 \cdot 42 + 5 \cdot 9.$

Continuing, we rearrange $(1)$ to substitute $9 = 93 - 2 \cdot 42$:

$3 = -1 \cdot 42 + 5 \cdot (93 - 2 \cdot 42)$

$3 = -11 \cdot 42 + 5 \cdot 93. \qquad (5)$

We have found one linear combination. To find others, since $42 \cdot 93 - 93 \cdot 42 = 0$, dividing both sides by $\gcd(93,42) = 3$ gives $14 \cdot 93 - 31 \cdot 42 = 0$. We can add $k$ times this equation to $(5)$, so we can write $3$ as a linear combination of $93$ and $42$

$3 = (14k + 5) \cdot 93 + (-31k - 11) \cdot 42$

for any integer $k$.

2016 AMC 10A Problems/Problem 25

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900$?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$

2018 AMC 10A Problems/Problem 22

Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$? (gcd means greatest common factor)

$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

2016 AMC 10B Problems/Problem 18

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

2017 AMC 10B Problems/Problem 14

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$?

$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$

Hint

2018 AMC 10B Problems/Problem 16

Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that\[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\]What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

2018 AMC 10A Problems/Problem 18

How many nonnegative integers can be written in the form

$a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0$, where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?

$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$