1. Euclid's Algorithm

Video coming soon2 worked examples

Theory

Euclid's algorithm finds the greatest common divisor (gcd) of two positive integers by repeated division. Write each step as $a = qb + r$ with $0 \leq r < b$ , then replace $(a, b)$ with $(b, r)$ and repeat until the remainder is $0$ . The last non-zero remainder is the gcd.

The gcd can then be written as a linear combination of the two numbers, $\gcd(a,b) = ax + by$ for integers $x, y$ , by back-substituting through the equations.

The Golden Rule: keep dividing until the remainder is $0$ ; the gcd is the last non-zero remainder. For the linear combination, work backwards through the algorithm's equations.

⚠️ Common Examiner Traps

Reading the wrong remainder: the gcd is the last non-zero remainder, not the final $0$ .
Back-substitution slips: substitute one equation at a time and keep the two original numbers visible so the coefficients stay correct.
Division form: each line must be $a = qb + r$ with $0 \leq r < b$ .

Worked examples

Example 1

Use Euclid's algorithm to find $\gcd(1071, 462)$ .

Step 1: Apply repeated division:

\begin{aligned} 1071 &= 2 \times 462 + 147 \\ 462 &= 3 \times 147 + 21 \\ 147 &= 7 \times 21 + 0 \end{aligned}

Step 2: The last non-zero remainder is $21$ , so $\gcd(1071, 462) = 21$ .

Example 2

Express $\gcd(1071, 462) = 21$ as a linear combination of $1071$ and $462$ .

Step 1: Rearrange the algorithm's equations to make the remainders the subject:

21 = 462 - 3 \times 147 \qquad 147 = 1071 - 2 \times 462

Step 2: Substitute the expression for $147$ and collect:

21 = 462 - 3(1071 - 2 \times 462) = 7 \times 462 - 3 \times 1071

Step 3: So $21 = (-3)(1071) + (7)(462)$ .