the ARML Quizlet: Part 1: Number Theory

Enumeration: Number Theory Problem Solving Techniques

"smart brute force" method →categories(classes, parity, values, etc.)

common pythagorean triples

(3, 4, 5) → 9+16=25 (5, 12, 13) → 25+144=169 (7, 24, 25) → 49+576=625 (8, 15, 17) → 64+225=289

Number Theory Problem Solving Techniques

1) Expressing Integers in Different Ways 2) Enumeration 3) Finding Patterns 4) Construction 5) Contradiction 6) Pairing 7) Estimation 8) Parity Analysis

lcm and gcd rules

1) If a|m and a|n, then a|gcd(m,n) 2) If m|k an n|k, then lcm(m,n)|k

modulo rules

1) If a≡b(mod m) and b≡c(mod m), then a≡c(mod m) 2) If a≡b(mod m) and c≡d(mod m), then (a+c)≡(b+d) (mod m) 3) If a≡b(mod m) and c≡d(mod m), then (ac)≡(bd) (mod m) 4) If a≡b(mod m) and k|m, then a≡b(mod k)

modular division rules

1) If a≡b(mod m), d|a, d|b, and gcd(d,m)=1, then (a/d)=(b/d)(mod m) 2) If a≡b(mod m), d|a, and d|b, then (a/d)=(b/d)(mod [m/gcd(d,m)])

Euclidean Algorithm

Assume m≥n and let r₀=m and r₁=n, then the there exists sequences of integers q₁....,qₙ and r₂....,rₙ such that: r₀=r₁q₁+r₂ with 0<r₂<r₁ r₁=r₂q₂+r₃ with 0<r₃<r₂ ... rₙ-₁=rₙqₙ with rₙ=gcd(m,n)

Division Algorithm

For any integers a and b, a≠0, there exists a unique pair (q,r) of integers such that b=aq+r and 0≤r<|a| → q is the quotient → r is the remainder

Bezout's Identity

For any two integers m and n, there exists integers a and b such that am+bn=gcd(m,n)

Wilson's Theorem

For every prime p, (p-1)!≡-1(mod p) Example: 2!=2≡−1(mod3)

modular multiplicative inverse

If and only if a and m are relatively prime, then there exists an integer b<m such that ba=1(mod m)

modular addition rule

If a≡b(mod m) and c≡d(mod m), then (a+c)≡(b+d) (mod m)

modular multiplication rule

If a≡b(mod m) and c≡d(mod m), then (ac)≡(bd) (mod m)

Fermat's Little Theorem

If p is prime and a is any integer, then p|([a^p]-a). Equivalently, if p does not divide a, then [a^p1-1]≡1(mod p)

Chinese Remainder Theorem

Let m₁....,mₙ be pairwise relatively prime positive integers (that is, gcd(m_i, m_j)=1 for all i≠j. Let b₁...,bₙ be arbitrary integers. Then the system- x=b₁ (mod m₁) x=b₂ (mod m₂) ... x=bₙ (mod mₙ) -has a unique solution modulo m₁m₂...mₙ

Π

Pi Notation; the product over a set of like terms (where Σ is the sum of a set of like terms)

residue classes

a complete set of integers that are congruent modulo m for some positive integer m; in modulo m, there are exactly n different residue classes, corresponding to the m possible residues {0,1,2,3,... m-2, m-1}

modular congruence

a mathematical way of saying that all of the integers are the same as one of the modulo m↔↔↔↔↔↔↔↔↔ residues

totative

a number less than or equal to and relatively prime to a given number

prime

a number p>1 if its divisors are itself and 1

primitive pythagorean triple

a pythagorean triple where gcd(a, b, c)=1 (the "base case" triples)

reduced set of residue class (system) modulo m

a set S of integers such that for each 0≤i≤m-1 where gcd(i,m)=1 there is an element sεS such that i≡s(mod m)

complete set of residue classes (system) modulo m

a set of integers S that for each 0≤i≤m-1, there is an element sεS such that i≡s(mod m)

pythagorean triple

a triple of positive integers (a, b, c) such that a²+b²=c²

primitive pythagorean triple generation

a=m²-n² b=2mn c=m²+n² where m, n are integers such that gcd(m,n)=1

the number a₁a₂a₃...aₙ in base 10 place value

a₁×10ⁿ⁻¹+a₂×10ⁿ⁻²+...+aₙ-₁×10+aₙ

Parity Analysis: Number Theory Problem Solving Techniques

categorize between even and odd →odd plus odd always equals odd

Construction: Number Theory Problem Solving Techniques

construct a set of numbers with special properties

modulus 16 power rule

every 4th power have remainders of 1 or 0

pythagorean triple generation

if a triple of integers (a, b, c) is a pythagorean triple, then so is (ka, kb, kc where k is any positive integer

factor

if an integer a divides an integer n evenly

divisible by 11

if a₀-a₁+a₂-...(-1)ⁿaₙ is divisibly by 11

divisible by 2

if divisible by 2

divisible by 5

if the number's last digit is a 0 or a 5

divisible by 3

if the sum of the digits is divisible by 3

divisible by 9

if the sum of the digits is divisible by 9

Contradiction: Number Theory Problem Solving Techniques

make an assumption opposite the desired conclusion, then use correct logical reasoning to reach a contradiction, and that proves that the original conclusion is correct

gcd(m,n)×lcm(m,n) =

mn

cototient of n

n-ϕ(n); gives the number of positive integers ≤n that have at least one prime factor in common with n

greatest common divisor

of m=p₁p₂p₃...pₙ an n=p₁p₂p₃....pₙ, is the largest number d such that d|m and d|n where d=Π{i=1,k} p(min)

least common multiple

of m=p₁p₂p₃...pₙ an n=p₁p₂p₃....pₙ, is the smallest number L such that m|L and n|L where L=Π{i=1,k} p(max)

Expressing Integers in Different Ways: Number Theory Problem Solving Techniques

place value technique n=a

Expressing Integers in Different Ways: Number Theory Problem Solving Techniques

place value technique n=aₙ×10ⁿ+aₙ-₁×10ⁿ⁻

Expressing Integers in Different Ways: Number Theory Problem Solving Techniques

place value technique n=aₙ×10ⁿ+aₙ-₁×10ⁿ⁻¹+...+a₁×10+a₀ division with a remainder n=mq+r where q is the quotient and r is the remainder prime factorization n=p₁p₂...pₙ power of 2 times an odd number m=2ⁿ×t where t is odd

Finding Patterns: Number Theory Problem Solving Techniques

solve similar problems with smaller sizes to find a pattern

divisible by 7 and 13

split the right-most 3 digits, subtract the smaller from the larger, check if the result is divisible by 7 and 13 (works for numbers with 5 and 6 digits)

modulus 3 and modulus 4 power rule

squares have remainders of 0 or 1

modulus 9 power rule

squares have remainders of 0, 1, 4, or 7

parity

the property of being even or odd

there are ___ possible remainders for modulo m

there are m possible remainders for modulo m

relatively prime

two integers m and n if gcd(m,n)=1

congruent modulo

two numbers a and b, if m|(a-b); if a and b have the same remainder when divided by m denoted a≡n(mod m) (a-b)/m for a≡b(mod m)

Estimation: Number Theory Problem Solving Techniques

use inequalities to shrink the range

prime Totient Function formula

where p is prime where q be some other prime dividing m

minimal complete set of residue classes

{0, 1,....,m-1}

reduced set of a prime number

{1, 2, ... p-1}

set of residue classes

{±1, ±2,...±k} where m=2k+1

number of factors

σ₀(n) = Π{i=1,k) [(e_i)+1] where (e_i) is the power of the prime factors

Euler Totient Function

ϕ(n); the number of elements in a reduced set of residue classes modulo m

prime Totient Function

ϕ(p)=p-1

convert to binary from base 10

→ for full numbers, divide by 2 and log the remainder → for decimal parts, multiply by 2 and take the integer part

Bezout's Identity [Proof]

→Let S={am+an|m,nεZ} →Where aa+bb=a²+b² so S contains positive elements →Prove t=(a,b) using t=av+bu →Division Algorithm a=tq+r →r=a-tq=a-(av+bu)q →so r<t but it must be positive so it is 0 so t|a and t|b →Prove t=c(ku+lv) using a=ck and b=cl →c≤t so so t is the gcd