A number which is formed by writing one digit $6$ times (e.g. $111111, 444444$ etc.), is always divisible by:

7

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11

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13

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17

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Solution

The correct options are
A $7$
B $13$
C $11$
$L e t t h e 6 d i g i t n u m b e r N b e r e p r e s e n t e d b y N = n \times 10^{6} + n \times 10^{5} + . . . n \times 10^{0} = n (10^{6} + 10^{5} + . . . . + 1) [n i s a n a t u r a l n u m b e r] = n (111111) N o w 111111 = 3 \times 7 \times 11 \times 13 \times 37 (c o m p l e t e l y f a c t o r i z e d) O p t i o n A ⟶ 7 i s a f a c t o r o f t h e n u m b e r . O p t i o n B ⟶ 11 i s a f a c t o r o f t h e n u m b e r . O p t i o n C ⟶ 13 i s a f a c t o r o f t h e n u m b e r . O p t i o n D ⟶ 17 i s n o t a f a c t o r o f t h e n u m b e r . ∴ N i s d i v i s i b l e b y 7, 11 a n d 13. A n s w e r - O p t i o n s A, B a n d C .$

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