Q. 75.0( 2 Votes )

Let abc be a thre A. a + b + c

B. 3

C. 37

D. 9

Answer :

abc = 100a + 10b + c.

bca = 100b + 10c + a.


cab = 100c + 10a + b.


abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b).


abc + bca + cab = 100a + 10a + a + 100b + 10b + b + 100c + 10c + c


abc + bca + cab = 111a + 111b + 111c.


abc + bca + cab = 111( a + b + c).


abc + bca + cab is divisible by 111 and (a + b + c) .


Also abc + bca + cab will be divisible by factors of 111


Option (a) abc + bca + cab is divisible by (a + b + c) because abc + bca + cab = 111(a + b + c).


Option (b) 3 is divisible by 111 because sum of digits of 111 is 3 and 3 is divisible by 3( Divisibility test of 3).


Option (c) 37 is the factor of 111 gives 0 as remainder and ( 37 x 4 = 111) therefore 111 is divisible by 37


Option (d) 9 is not factor of 111 because not gives 0 as remainder. Therefore 111 is not divisible by 9 .

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