GCD and LCM: method and worked exercises

It is the first days of October: in Year 1 of high school the maths teacher is reviewing fractions and powers, and in three to five weeks the first test arrives, the one with expressions full of fractions. Whoever gets those expressions wrong, almost always, does not get the fraction wrong: they get the common denominator wrong, that is the LCM (lowest common multiple). And whoever cannot factorise into prime factors finds it even harder in January, when factorisation arrives with letters.

This post is about the piece that lies under everything: factorising a number, calculating the GCD (greatest common divisor) and LCM, using them. If you get there in time, the fractions in the test become a matter of order.

Divisibility rules: the five you need

To factorise you need to know what a number is divisible by, without doing the division. These are enough:

by $2$ : the last digit is even;
by $3$ : the sum of the digits is a multiple of $3$ (example: $252 \to 2 + 5 + 2 = 9$ , so yes);
by $5$ : it ends in $0$ or $5$ ;
by $9$ : the sum of the digits is a multiple of $9$ ;
by $11$ : the sum of the digits in even places and that of the digits in odd places differ by a multiple of $11$ (used less, but it happens).

For $4$ look at the last two digits: if they form a multiple of $4$ (such as $12$ , $36$ , $80$ ), so is the whole number.

Prime factorisation

A prime number has only two divisors: $1$ and itself ( $2, 3, 5, 7, 11, 13 \dots$ ). Factorising a number means writing it as a product of primes, raised to a power when they repeat. You proceed by dividing by the smallest possible prime, until you reach $1$ :

360 = 2 \cdot 180 = 2 \cdot 2 \cdot 90 = 2 \cdot 2 \cdot 2 \cdot 45 = 2^{3} \cdot 3 \cdot 15 = 2^{3} \cdot 3^{2} \cdot 5

The check is immediate: $8 \cdot 9 \cdot 5 = 360$ . I always multiply back, because a wrong factorisation ruins everything that follows.

Another example: $252 = 2^{2} \cdot 3^{2} \cdot 7$ , because $4 \cdot 9 \cdot 7 = 252$ .

GCD: the greatest common divisor

The GCD of two numbers is the largest number that divides both. It is calculated like this: you take only the common factors, each with the smaller exponent.

With $360 = 2^{3} \cdot 3^{2} \cdot 5$ and $252 = 2^{2} \cdot 3^{2} \cdot 7$ :

the $2$ is in both: exponents $3$ and $2$ , you take $2^{2}$ ;
the $3$ is in both: exponents $2$ and $2$ , you take $3^{2}$ ;
the $5$ and the $7$ appear in only one, you do not take them.

GCD (360, 252) = 2^{2} \cdot 3^{2} = 36

LCM: the lowest common multiple

The LCM is the smallest number that is a multiple of both. Here you take all the factors, common and non-common, each with the larger exponent:

LCM (360, 252) = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7 = 2520

A check I always do: the product of the two numbers equals the product of the GCD and the LCM. Here $360 \cdot 252 = 90, 720$ and $36 \cdot 2520 = 90, 720$ . It works only for two numbers at a time, but for checking it is perfect.

To remember which goes with the minimum and which with the maximum: the GCD looks for what lies inside both, so it is cautious (minimum, only the common ones); the LCM must contain both, so it is generous (maximum, all of them).

What they are for: two uses

Adding and subtracting fractions

To add fractions you need a common denominator, and the most convenient is the LCM of the denominators. Example:

\frac{5}{12} + \frac{7}{18}

You factorise: $12 = 2^{2} \cdot 3$ and $18 = 2 \cdot 3^{2}$ , so $LCM = 2^{2} \cdot 3^{2} = 36$ . Then you convert each fraction: $36 : 12 = 3$ and $36 : 18 = 2$ .

\frac{5 \cdot 3}{36} + \frac{7 \cdot 2}{36} = \frac{15 + 14}{36} = \frac{29}{36}

$29$ is prime, so it does not simplify: the fraction is done. Use the LCM and not just any common denominator ( $12 \cdot 18 = 216$ works, but the numbers grow and you make more mistakes).

Simplifying with the GCD

To reduce to lowest terms you divide numerator and denominator by their GCD:

\frac{84}{126} : 84 = 2^{2} \cdot 3 \cdot 7, 126 = 2 \cdot 3^{2} \cdot 7, GCD = 2 \cdot 3 \cdot 7 = 42

\frac{84}{126} = \frac{84 : 42}{126 : 42} = \frac{2}{3}

Word problems

Problems are of two types and it is worth recognising them from the text:

“How often do they meet / when do they set off together”: LCM. You are looking for a future moment when two periodic things coincide.
“The largest measure / the largest piece that fits a whole number of times”: GCD. You are looking for a piece that divides two lengths exactly.

An example of the second type. A floor is $240 cm$ long and $180 cm$ wide. You want square tiles, as large as possible, without cutting any. The side must divide both $240$ and $180$ , and be the maximum: it is the GCD. With $240 = 2^{4} \cdot 3 \cdot 5$ and $180 = 2^{2} \cdot 3^{2} \cdot 5$ we get $GCD = 2^{2} \cdot 3 \cdot 5 = 60$ . Side $60 cm$ : along the length $240 : 60 = 4$ fit, along the width $180 : 60 = 3$ , in all $4 \cdot 3 = 12$ tiles.

Three worked exercises

Exercise 1: GCD, LCM and a reduced fraction

Calculate the GCD and LCM of $168$ and $270$ and use them to reduce $\frac{168}{270}$ to lowest terms.

Factorisations: $168 = 2^{3} \cdot 3 \cdot 7$ (check: $8 \cdot 3 \cdot 7 = 168$ ) and $270 = 2 \cdot 3^{3} \cdot 5$ (check: $2 \cdot 27 \cdot 5 = 270$ ).

Common factors: $2$ (exponents $3$ and $1$ , minimum $2^{1}$ ) and $3$ (exponents $1$ and $3$ , minimum $3^{1}$ ).

GCD = 2 \cdot 3 = 6 LCM = 2^{3} \cdot 3^{3} \cdot 5 \cdot 7 = 8 \cdot 27 \cdot 35 = 7560

Check: $168 \cdot 270 = 45, 360$ and $6 \cdot 7560 = 45, 360$ . It works. The fraction:

\frac{168}{270} = \frac{168 : 6}{270 : 6} = \frac{28}{45}

If you want to try more calculations of this type against the clock, you will find the simulated test on numbers, fractions and powers: it covers the October block, where GCD and LCM are one piece, not the only topic. The correct answer to every question is free, the guided step-by-step solution is unlocked with the first lesson.

Exercise 2: an expression with three fractions

\frac{7}{15} - \frac{3}{20} + \frac{1}{12}

Factorisation of the denominators: $15 = 3 \cdot 5$ , $20 = 2^{2} \cdot 5$ , $12 = 2^{2} \cdot 3$ . The LCM takes all the factors with the larger exponent: $2^{2} \cdot 3 \cdot 5 = 60$ .

Conversion factors: $60 : 15 = 4$ , $60 : 20 = 3$ , $60 : 12 = 5$ .

\frac{7 \cdot 4}{60} - \frac{3 \cdot 3}{60} + \frac{1 \cdot 5}{60} = \frac{28 - 9 + 5}{60} = \frac{24}{60}

It is reduced with $GCD (24, 60)$ : $24 = 2^{3} \cdot 3$ , $60 = 2^{2} \cdot 3 \cdot 5$ , so $GCD = 2^{2} \cdot 3 = 12$ .

\frac{24}{60} = \frac{2}{5}

Exercise 3: two buses

Two buses set off together from the terminus at 7:00. The first comes back every $12$ minutes, the second every $18$ . At what time do they find themselves together again?

It is an LCM problem: $12 = 2^{2} \cdot 3$ and $18 = 2 \cdot 3^{2}$ , so $LCM = 2^{2} \cdot 3^{2} = 36$ . They meet again after $36$ minutes, that is at 7:36. In those $36$ minutes the first has made $36 : 12 = 3$ runs, the second $36 : 18 = 2$ . The check with whole runs confirms that $36$ is the first common instant.

The mistakes I see most

Swapping minimum and maximum. Using the larger exponent for the GCD or the smaller for the LCM. A quick test: the GCD must be less than or equal to the smaller of the two numbers; the LCM must be greater than or equal to the larger.
Not factorising all the way. Stopping at $360 = 6 \cdot 60$ is not a prime factorisation: $6$ and $60$ are not prime.
Taking only the common factors in the LCM. The LCM also includes those that are in only one number. In exercise 2 the $5$ is in two denominators, but the $2^{2}$ and the $3$ must be taken anyway.
Adding the denominators. $\frac{1}{3} + \frac{1}{4}$ is not $\frac{2}{7}$ . The common denominator is there precisely so that you do not do that.
Forgetting to multiply the numerator. You convert the denominator with the factor $60 : 15 = 4$ and you also multiply the numerator by $4$ . It is one of the commonest mistakes in the test.

Why now and not in January

In January, in algebra, polynomials are factorised by taking out the common factor, and the procedure is the same with letters: for $12 a^{2} b$ and $18 a b^{3}$ we have $GCD = 6 ab$ and $LCM = 36 a^{2} b^{3}$ (the same rules, minimum and maximum of the exponents for letters too). Anyone who is confident with numbers arrives already prepared. Calculation with letters, monomials and polynomials is in algebraic calculation: monomials and polynomials.

For the October test a simple plan pays off: every day factorise five random numbers between $50$ and $500$ , then do two sums of fractions with three different denominators. After a week factorisation becomes automatic. The full picture of what comes up in the first test is in the first maths test in Year 1 of high school; if maths seems harder than expected and you want to restart from numbers, there is numbers, fractions and powers (written for the art high school, but the method holds everywhere); for proportions and percentages, which come straight after, proportions and percentages in Year 1.

If you want to see where to restart, the free introductory call is for exactly this: you send me the marked test or tell me what you have done in class and I tell you whether it is enough to fix the method or whether it pays to work on it together. The maths programme for the first two years is here; lessons are online and one-to-one.

GCD and LCM: how to calculate them and when to use them