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