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Question

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


A
7
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B
11
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C
13
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D
17
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Solution

The correct options are
A $$7$$
B $$13$$
C $$11$$

$$ Let\  the\  6\  digit\  number\  N\  be\  represented\  by\  N=n\times { 10 }^{ 6 }+n\times { 10 }^{ 5 }+...n\times { 10 }^{ 0 }\\ \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  =n\left( { 10 }^{ 6 }+{ 10 }^{ 5 }+....+1 \right) \left[ n\  is\  a\  natural\  number \right] \\ \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  \  =n\left( 111111 \right) \\ Now\  111111=3\times 7\times 11\times 13\times 37\  (completely\  factorized)\\ Option\  A\longrightarrow 7\  is\  a\  factor\  of\  the\  number.\\ Option\  B\longrightarrow 11\  is\  a\  factor\  of\  the\  number.\\ Option\  C\longrightarrow 13\  is\  a\  factor\  of\  the\  number.\\ Option\  D\longrightarrow 17\  is\  not\  a\  factor\  of\  the\  number.\\ \therefore \  N\  is\  divisible\  by\  7,\  11\  and\  13.\\ Answer-\  Options\  A,\  B\  and\  C. $$


Mathematics

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