Question

Find the smallest number which when divided by 35, 56 and 105 leaves a remainder of 6 in each case.

Answer 1

Answer:

Step-by-step explanation:

Let the number be N. The number N can be written in the following ways.

N = 35x + 6 —(1)

N = 56y + 6 —(2)

N = 105z + 6 —(3)

where x,y and z are whole numbers.

Substracting eqn(1) from eqn(2) we get,

0 = 56y-35x,

i.e 56y = 35x,

i.e 8y = 5x —(4)

Similarly subtracting eqn(1) from eqn(3), we get,

0 = 105z - 35x,

i.e 35x = 105z

x = 3z —(5)

The minimum values of x, y and z satisfying eqn (4) and (5) by trial and error method is x=24, y=15 and z=8.

Substituting z in eqn(3), we get,

N = 105*8 + 6

N = 846