Question

find the HCF of the following factorisation METHOD: problem 315,540?

Answer 1

Approach 1. Integer numbers prime factorization:

315 = 32 × 5 × 7; 
540 = 22 × 33 × 5;

Take all the common prime factors, by the lowest exponents.

Greatest (highest) common factor (divisor)
gcf, gcd (315; 540) = 32 × 5 = 45;

Approach 2. Euclid's algorithm:

Step 1. Divide the larger number by the smaller one:
540 ÷ 315 = 1 + 225;

Step 2. Divide the smaller number by the above operation's remainder: 
315 ÷ 225 = 1 + 90;Step 3. Divide the remainder from the step 1 by the remainder from the step 2: 
225 ÷ 90 = 2 + 45;Step 4. Divide the remainder from the step 2 by the remainder from the step 3: 
90 ÷ 45 = 2 + 0;At this step, the remainder is zero, so we stop: 
45 is the number we were looking for, the last remainder that is not zero. 
This is the greatest common factor (divisor).

Greatest (highest) common factor (divisor)
gcf, gcd (315; 540) = 45 = 32 × 5;

Answer 2

315 = 3 x 3 x 5 x 7
540 = 2 x 2 x 3 x 3 x 3 x 5
so, the hcf = 3 x 3 x 5 = 45