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Bunuel
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Read the question:

aXXa is divisible by 6, i.e divisible by 2 and 3

only 6126 is divisible by 6.
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Solution:

a12a is divisible by 6 means it has to be divisible by 2 and 3

To be divisible be 2, the last digit i.e., a has to be even. This eliminates options B and E

Now we can plug each option and check divisibility by 3

Option A: 2
a12a = 2122 is not divisible by 3 because 2 + 1 + 2 + 2 = 7 is not divisible by 3

Option C: 4
a12a = 4124 is not divisible by 3 because 4 + 1 + 2 + 4 = 11 is not divisible by 3

Option D: 6
a12a = 6126 is divisible by 3 because 6 + 1 + 2 + 6 = 15 is divisible by 3

Hence the right answer is Option D
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