Question

When 7897,8110,8536 are divided by number x,the remainder in each case is same.find x

Answer 1

Let the remainder be r in each case. Thus, let,

7897 = ax + r

7897 - r = ax → (1)

$\quad$

8110 = bx + r

8110 - r = bx → (2)

$\quad$

8536 = cx + r

8536 - r = cx → (3)

where a,b and c are the respectively quotients.

$\quad$

Subtracting (1) from (2),

(8110 - r) - (7897 - r) = bx - ax

213 = (b - a)x

This means 213 is a multiple of x.

$\quad$

Subtracting (2) from (3),

(8536 - r) - (8110 - r) = cx - bx

426 = (c - b)x

This means 426 is a multiple of x.

$\quad$

Subtracting (1) from (3),

(8536 - r) - (7897 - r) = cx - ax

639 = (c - a)x

This means 639 is a multiple of x.

$\quad$

Then we can say that x is a common factor of the numbers 213, 426 and 639.

$\quad$

Then the following set X contains the possible values of x.

X = {1, 3, 71, 213}