The sum of all numbers formed by the digits x,y and z of the number xyz is divisible by

Question

Philoid · Accepted Answer

Three numbers can be formed by the digits x,y and z they are xyz, yzx and zxy.

⇒ xyz = 100x + 10y + z.

⇒ yzx = 100y + 10z + x.

⇒ zxy = 100z + 10x + y.

⇒ xyz + yzx + zxy = (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y).

⇒ xyz + yzx + zxy = 100x + 10x + x + 100y + 10y + y + 100z + 10z + z

⇒ xyz + yzx + zxy = 111x + 111y + 111z.

⇒ xyz + yzx + zxy = 111( x + y + z).

⇒ xyz + yzx + zxy is divisible by 111 because 111 is the common factor .

Also xyz + yzx + zxy will be divisible by factors of 111.

Option (a) ⇒ 11 is not the factor of 111 (becausedoes not give 0 as remainder).

Option (b) ⇒ 33 is not the factor of 111 (becausedoes not give 0 as remainder).

Option (c) ⇒ 37 is the factor of 111 gives 0 as remainder and ( 37 x 4 = 111). Therefore it is correct option.

Option (d) ⇒ 74 is not the factor of 111 (becausedoes not give 0 as remainder).