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The product rule, and reading factorisations

Class 10Maths HCF times LCM, and arguing from primesSession 3 Free

For two positive whole numbers, HCF times LCM equals the product of the numbers. Prime factorisations also settle questions about divisibility for every case at once, without trial.

A relation between HCF and LCM

For 6 and 20, the HCF is 2 and the LCM is 60. Their product is 2 × 60 = 120. The product of the two numbers is 6 × 20 = 120. The two results are equal.

This holds for any two positive whole numbers a and b:

HCF(a, b) × LCM(a, b) = a × b

Why does HCF × LCM equal a × b? Follow one prime through the calculation.

Take any prime p. Suppose a contains p three times and b contains it five times. Then a × b contains p a total of 3 + 5 = 8 times.

The HCF takes the smaller count, 3. The LCM takes the larger count, 5. So HCF × LCM contains p a total of 3 + 5 = 8 times.

For any two counts, smaller + larger = first + second. This is true for every prime, so HCF × LCM and a × b have identical prime factorisations. By uniqueness, they are the same number.

Using the relation

If the HCF is known, the LCM follows from one division:

LCM(a, b) = (a × b) ÷ HCF(a, b)

Given HCF(306, 657) = 9: LCM = (306 × 657) ÷ 9 = 201,042 ÷ 9 = 22,338.

It applies to two numbers only

For three numbers, HCF × LCM is generally not equal to their product. For example, with 2, 4 and 8: HCF = 2, LCM = 8, so HCF × LCM = 16, but 2 × 4 × 8 = 64. The argument above needed two counts per prime; with three counts, smallest + largest leaves out the middle one.

Arguing from prime factorisations

A prime factorisation can decide a question for infinitely many cases in a few lines. The previous sessions showed that 4ⁿ never ends in the digit 0. The same method works for other numbers.

Can 6ⁿ end with the digit 0, for any natural number n?

A number ends in 0 exactly when it is divisible by 10 = 2 × 5. So its prime factorisation must contain both 2 and 5.

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. Its only prime factors are 2 and 3. It contains no 5.

By uniqueness of prime factorisation, 6ⁿ has no other factorisation that could contain 5. So 6ⁿ never ends in 0, for any n.

Showing a number is composite without computing it

Consider 7 × 11 × 13 + 13. Both terms contain the factor 13. Taking it out gives 13 × (7 × 11 + 1) = 13 × 78.

The number is a product of two whole numbers, 13 and 78, each greater than 1. So it has a factor other than 1 and itself, and it is composite. Its value, 1014, never needed to be calculated.

Check what you have understood

1For any two positive whole numbers a and b, HCF(a, b) × LCM(a, b) = ____.

a × b. For each prime, the smaller count plus the larger count equals the two counts added together.

2Given HCF(306, 657) = 9, the LCM is…

LCM = (a × b) ÷ HCF = 201,042 ÷ 9 = 22,338.

3The rule HCF × LCM = product holds for…

For two numbers. With 2, 4 and 8: HCF × LCM = 16, but the product is 64.

4Why can 6ⁿ never end in the digit 0?

6ⁿ contains only the primes 2 and 3. By uniqueness it can never contain 5.

5Why is 7 × 11 × 13 + 13 composite?

A factor other than 1 and itself, here 13, makes it composite.

6A prime appears four times in a and once in b. How many times does it appear in HCF × LCM?

HCF: 1 time. LCM: 4 times. Together 5, the same as 4 + 1 in a × b.

Next: how the primes of a denominator decide whether a decimal terminates or repeats.

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