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Q.

Let the three digit numbers A28, 3B9, 62C  where A, B, C are integers between 0 and 9 be divisible by a fixed integer K then A3689C2B2 is divisible by

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a

K

b

K+2

c

K2

d

KK+1

answer is C.

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Detailed Solution

Let A28=ka, 3B9=kb, 62C=kc be divisible by 'k' when a,b,c an integers.

      apply R2R2+100R1+10R3

A36A283B962C2B2=kA36abc2B2

       = integral multiple of k

         is divisible by 'k'                    

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