Number TheoryTopic #30 of 40

Divisibility

Division algorithm, divisibility rules, GCD, LCM, and the Euclidean algorithm.

Definition

An integer aa divides an integer bb (written aba \mid b) if there exists an integer kk such that: b=akb = ak

We say aa is a divisor or factor of bb, and bb is a multiple of aa.

If aba \nmid b, then aa does not divide bb.

Basic Properties

For all integers a,b,ca, b, c:

  1. a0a \mid 0 (every integer divides 0)
  2. 1a1 \mid a (1 divides every integer)
  3. If aba \mid b and bcb \mid c, then aca \mid c (transitivity)
  4. If aba \mid b and aca \mid c, then a(b+c)a \mid (b + c) and a(bc)a \mid (b - c)
  5. If aba \mid b, then abca \mid bc for any integer cc
  6. If aba \mid b and aca \mid c, then a(mb+nc)a \mid (mb + nc) for any integers m,nm, n

For positive integers:

If aba \mid b and b0b \neq 0, then ab|a| \leq |b|.

Division Algorithm

For any integers aa and dd with d>0d > 0, there exist unique integers qq (quotient) and rr (remainder) such that: a=dq+rwhere 0r<da = dq + r \quad \text{where } 0 \leq r < d

Notation

  • a÷d=qa \div d = q (integer division)
  • amodd=ra \mod d = r (remainder)

Examples

  • 17=5×3+217 = 5 \times 3 + 2, so 17÷5=317 \div 5 = 3 and 17mod5=217 \mod 5 = 2
  • 17=5×(4)+3-17 = 5 \times (-4) + 3, so 17÷5=4-17 \div 5 = -4 and 17mod5=3-17 \mod 5 = 3

Greatest Common Divisor (GCD)

The greatest common divisor of aa and bb, written gcd(a,b)\gcd(a, b) or (a,b)(a, b), is the largest positive integer that divides both.

Properties

gcd(a,b)=gcd(b,a)\gcd(a, b) = \gcd(b, a) gcd(a,0)=a\gcd(a, 0) = |a| gcd(a,b)=gcd(ab,b)=gcd(amodb,b)\gcd(a, b) = \gcd(a - b, b) = \gcd(a \mod b, b) gcd(a,b)=gcd(a,ba)\gcd(a, b) = \gcd(a, b - a)

Coprime (Relatively Prime)

aa and bb are coprime if gcd(a,b)=1\gcd(a, b) = 1.

Euclidean Algorithm

Efficiently computes gcd(a,b)\gcd(a, b) using repeated division.

Algorithm

gcd(a, b):
    while b ≠ 0:
        r = a mod b
        a = b
        b = r
    return a

Example

gcd(252,105)\gcd(252, 105) 252=105×2+42252 = 105 \times 2 + 42 105=42×2+21105 = 42 \times 2 + 21 42=21×2+042 = 21 \times 2 + 0

gcd(252,105)=21\gcd(252, 105) = 21

Time Complexity

O(log(min(a,b)))O(\log(\min(a, b))) divisions

Extended Euclidean Algorithm

Finds integers xx and yy such that: gcd(a,b)=ax+by\gcd(a, b) = ax + by

Bézout's Identity

For any integers aa and bb (not both zero), there exist integers xx and yy such that: gcd(a,b)=ax+by\gcd(a, b) = ax + by

Example

Find x,yx, y such that gcd(252,105)=252x+105y\gcd(252, 105) = 252x + 105y

Working backwards: 21=10542×221 = 105 - 42 \times 2 21=105(252105×2)×221 = 105 - (252 - 105 \times 2) \times 2 21=105×5252×221 = 105 \times 5 - 252 \times 2 21=252×(2)+105×521 = 252 \times (-2) + 105 \times 5

So x=2x = -2, y=5y = 5.

Least Common Multiple (LCM)

The least common multiple of aa and bb, written lcm(a,b)\text{lcm}(a, b) or [a,b][a, b], is the smallest positive integer divisible by both.

Relationship to GCD

gcd(a,b)×lcm(a,b)=ab\gcd(a, b) \times \text{lcm}(a, b) = |ab|

lcm(a,b)=abgcd(a,b)\text{lcm}(a, b) = \frac{|ab|}{\gcd(a, b)}

Properties

lcm(a,b)=lcm(b,a)\text{lcm}(a, b) = \text{lcm}(b, a) lcm(a,1)=a\text{lcm}(a, 1) = |a| lcm(a,a)=a\text{lcm}(a, a) = |a|

Divisibility Rules

DivisorRule
2Last digit is even
3Sum of digits divisible by 3
4Last two digits form number divisible by 4
5Last digit is 0 or 5
6Divisible by both 2 and 3
8Last three digits divisible by 8
9Sum of digits divisible by 9
10Last digit is 0
11Alternating sum of digits divisible by 11

Linear Diophantine Equations

Equation of the form: ax+by=cax + by = c

Solvability

Has integer solutions if and only if gcd(a,b)c\gcd(a, b) \mid c.

Finding Solutions

  1. Find particular solution (x0,y0)(x_0, y_0) using extended Euclidean algorithm
  2. General solution: x=x0+bdt,y=y0adtx = x_0 + \frac{b}{d}t, \quad y = y_0 - \frac{a}{d}t where d=gcd(a,b)d = \gcd(a, b) and tZt \in \mathbb{Z}.