Q. 55.0( 2 Votes )

# The sum of all nu

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 (because does not give 0 as remainder).

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

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