Question

Find the greatest number that will divide $84,103$ and $122$ leaving remainder $4,3$ and $2$ respectively.

Answer 1

Step 1 Subtract the remainders that are $\mathbf{4}\mathbf{,}\mathbf{}\mathbf{3}$ and $\mathbf{2}$ from the given numbers $\mathbf{84}\mathbf{,}\mathbf{}\mathbf{103}$ and $\mathbf{122}$ respectively :

$84-4=80$

$103-3=100$

$122-2=120$

Step 2 Finding highest common factor of the numbers obtained in step 1 ( by subtracting the remainder from the given numbers) that are $\mathbf{80}\mathbf{,}\mathbf{}\mathbf{100}$and $\mathbf{120}$:

Highest common factor of $\mathbf{80}\mathbf{,}\mathbf{}\mathbf{100}$and $\mathbf{120}$ (by prime factorisation method ) :

$80=2\times 2\times 2\times 2\times 5=2^4\times 5$

$100=2\times 2\times 5\times 5=2^2\times 5^2$

$120=2\times 2\times 2\times 5\times 3=2^3\times 3\times 5$

So Highest common factor of $80,100$ and $120$ $=2^2\times 5$ ( $\because$ multiplying the common prime factors to get highest common factors )

$=20$

Since the greatest number that will divide $84,103$ and $122$ leaving the remainder $4,3$ and $2$ is highest common factor of $80,100$ and $120$

Therefore Greatest number that will divide $\mathbf{84}\mathbf{,}\mathbf{}\mathbf{103}$ and $\mathbf{122}$leaving remainder $\mathbf{4}\mathbf{,}\mathbf{}\mathbf{3}$ and $\mathbf{2}$ respectively is $\mathbf{20}$