M303 20260418 Number Theory

1962 AHSME Problems/Problem 21

It is given that one root of $2x^2 + rx + s = 0$, with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$. The value of $s$ is:

$\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ -13\qquad \textbf{(E)}\ 26$

#The_Complex_Conjugate_Root_Theorem

If $p(x)$ is a polynomial with real coefficients, and $z$ is a complex root of $p(x)$, then its complex conjugate $\bar{z}$ is also a root of $p(x)$.

Key Mathematical Properties Used

The proof relies on three fundamental properties of complex conjugates:

1. Distributivity over Addition: The conjugate of a sum is the sum of the conjugates.

$$\overline{z + w} = \bar{z} + \bar{w}$$

2. Distributivity over Multiplication: The conjugate of a product is the product of the conjugates.

$$\overline{zw} = \bar{z} \bar{w}$$

3. Polynomial Application: For a polynomial $p(x)$ with real coefficients, the conjugate of the polynomial evaluated at $z$ is equal to the polynomial evaluated at the conjugate of $z$.

$$\overline{p(z)} = p(\bar{z})$$

If $n$ is a multiple of 4, the sum $s = 1 + 2i + 3i^2 + \dots + (n + 1)i^n$, where $i = \sqrt{-1}$, equals:

(A) $1 + i$ (B) $\frac{1}{2}(n + 2)$ (C) $\frac{1}{2}(n + 2 - ni)$ (D) $\frac{1}{2}[(n + 1)(1 - i) + 2]$ (E) $\frac{1}{8}(n^2 + 8 - 4ni)$

Calculus Solution:

To differentiate a fraction, you use the Quotient Rule.

If you have a function $f(x) = \frac{u}{v}$, the derivative is:

$$f'(x) = \frac{u'v - uv'}{v^2}$$

The Steps:

1. Differentiate the top ($u'$) and the bottom ($v'$) separately.

2. Apply the formula: (Top's derivative $\times$ Bottom) minus (Top $\times$ Bottom's derivative).

3. Divide the whole thing by the Bottom squared.

Simple Memory Trick:

> "Low D-High minus High D-Low, over Low-Low"

Low = Bottom ($v$)
High = Top ($u$)
D = Derivative

Example from your image:

For $p(x) = \frac{1-x^{n+2}}{1-x}$:

Top ($u$): $1-x^{n+2} \rightarrow u' = -(n+2)x^{n+1}$
Bottom ($v$): $1-x \rightarrow v' = -1$

$$p'(x) = \frac{\overbrace{[-(n+2)x^{n+1}]}^{u'} \cdot \overbrace{(1-x)}^{v} - \overbrace{(1-x^{n+2})}^{u} \cdot \overbrace{(-1)}^{v'}}{\underbrace{(1-x)^2}_{v^2}}$$

$$p'(x) = \frac{-(n+2)x^{n+1}(1-x) + (1-x^{n+2})}{(1-x)^2}$$

The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^2$? Note: if $z = a + bi$, then $|z| = \sqrt{a^2 + b^2}$.

(A) 68 (B) 100 (C) 169 (D) 208 (E) 289

1987 AHSME Problems/Problem 28

Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle in the complex plane centered at $0+0i$ and having radius $1$. The sum of the reciprocals of the roots is necessarily

$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$

1992 AHSME Problems/Problem 28

Let $i = \sqrt{-1}$. The product of the real parts of the roots of $z^2 - z = 5 - 5i$ is

(A) $-25$ (B) $-6$ (C) $-5$ (D) $\frac{1}{4}$ (E) 25

2002 AMC 12A Problems/Problem 24

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$.

$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

2007 AMC 12A Problems/Problem 11

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$

2009 AMC 10A Problems/Problem 25

For $k > 0$, let $I_k = 10\ldots 064$, where there are $k$ zeros between the $1$ and the $6$. Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$. What is the maximum value of $N(k)$?

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

#Modular_Congruence

Two numbers $a$ and $b$ are congruent modulo $m$ if their difference $(a - b)$ is a multiple of $m$.

$$a \equiv b \pmod{m}$$

Examples provided:

$1 \equiv 8 \equiv 15 \equiv 22 \pmod{7}$ (Each differs by a multiple of 7)
$2 \equiv 14 \equiv 26 \equiv 38 \pmod{12}$ (Each differs by a multiple of 12)

Properties

If we suppose $a \equiv b \pmod{m}$ and $c \equiv d \pmod{m}$, then the following arithmetic rules apply:

1. Addition:

$$a + c \equiv b + d \pmod{m}$$

2. Subtraction:

$$a - c \equiv b - d \pmod{m}$$

3. Multiplication:

$$ac \equiv bd \pmod{m}$$

4. Power (Exponentiation):

$$a^n \equiv b^n \pmod{m}$$

2009 AMC 10B Problems/Problem 21

What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?

$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$

Solution Strategy

$a \equiv b \pmod{m}$

1. look for a pattern

2. #Eulers_Totient_Theorem

3. #Binomial_Theorem

Suppose $\gcd(a, n) = 1$ (meaning $a$ and $n$ are relatively prime). Then:

$$a^{\phi(n)} \equiv 1 \pmod{n}$$

Euler's Totient Function: $\phi(n)$

Definition: $\phi(n)$ is the number of integers $\le n$ that are relatively prime to $n$.
Formula:

$$\phi(n) = n \cdot \prod_{p|n} \left( 1 - \frac{1}{p} \right)$$

(Where $p$ represents the distinct prime factors of $n$).

Example: $n = 100$

The prime factors of $100$ are $2$ and $5$ (since $100 = 2^2 \cdot 5^2$).

1. Calculate $\phi(100)$:

$$\phi(100) = 100 \cdot \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{5} \right)$$

$$\phi(100) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40$$

2. Apply the Theorem:

If $\gcd(a, 100) = 1$, then:

$$a^{40} \equiv 1 \pmod{100}$$

Question:

a) Please find $\phi(1000)$.

b) Find $3^{402} \pmod{1000}$.

2011 AMC 10B Problems/Problem 23

What is the hundreds digit of $2011^{2011}?$

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

How base works?

Positional Notation

A number is the sum of its digits multiplied by the base raised to the power of their position (starting from $0$ at the right).

Base 10 (Decimal):

$$12345_{10} = (1 \cdot 10^4) + (2 \cdot 10^3) + (3 \cdot 10^2) + (4 \cdot 10^1) + (5 \cdot 10^0)$$

Base 6 (Senary):

$$12345_{6} = (1 \cdot 6^4) + (2 \cdot 6^3) + (3 \cdot 6^2) + (4 \cdot 6^1) + (5 \cdot 6^0)$$

How to convert from base to base?

From base 6 to base 10:

Calculate the powers of 6:

$6^4 = 1296$
$6^3 = 216$
$6^2 = 36$
$6^1 = 6$
$6^0 = 1$

Total: $1296 + 432 + 108 + 24 + 5 = \mathbf{1865_{10}}$

From base 10 to base 6:

To convert from Base 10 to Base 6, you use the Successive Division Method. You repeatedly divide the number by 6 and keep track of the remainders.

The Steps

1. Divide the decimal number by 6.

2. Write down the remainder (this will be between 0 and 5).

3. Take the quotient (the whole number result) and divide it by 6 again.

4. Repeat until the quotient is 0.

5. Read the remainders in reverse order (from the last one found to the first).

Division	Quotient	Remainder
$1865 \div 6$	310	5 (First digit from right)
$310 \div 6$	51	4
$51 \div 6$	8	3
$8 \div 6$	1	2
$1 \div 6$	0	1 (Last digit from left)

2013 AMC 10A Problems/Problem 19

In base $10$, the number $2013$ ends in the digit $3$. In base $9$, on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$. For how many positive integers $b$ does the base-$b$-representation of $2013$ end in the digit $3$?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Based on the definition of base systems, the units digit (last digit) of any number $N$ in base $b$ is simply the remainder of $N$ divided by $b$.

根据进位制的定义，任何数 $N$ 在 $b$ 进制下的个位数，其实就是 $N$ 除以 $b$ 的余数。

The problem asks for the number of positive integers $b$ such that the base-$b$ representation of $2013$ ends in $3$.

Here is the breakdown:

The Logic: If $2013$ ends in $3$ in base $b$, it means that when $2013$ is divided by $b$, the remainder is $3$.
The Equation: $2013 \equiv 3 \pmod{b}$, which simplifies to $2013 - 3 = 2010$. Therefore, $b$ must be a factor of $2010$.
Finding Factors: Prime factorize $2010$:

$$2010 = 2 \times 3 \times 5 \times 67$$

The total number of factors is $(1+1)(1+1)(1+1)(1+1) = 16$.

The Constraint: In any base $b$, the digits must be smaller than $b$. Since the last digit is $3$, the base $b$ must be greater than $3$ ($b > 3$).
Final Count: We exclude the factors of $2010$ that are $\le 3$. Those are $1, 2,$ and $3$.

$$16 - 3 = 13$$

The answer is 13.

2015 AMC 10A Problems/Problem 18

Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$. Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$?

$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$

M203 课程目录