Let f(1) = f(4) = A ; f(2) = f(5) = B ; f(3) = f(6) = C then,

A. (10^5 * C ) + ( 10^4 *B) + (10^3 *A ) + ( 10^2 * C ) + (10 * B) + A

Reducing the terms we have

( 10^2 * C ) * [ 10^3 + 1 ] + (10 * B) * [ 10^3 + 1 ] + A * [ 10^3 + 1 ]

Taking [ 10^3 + 1 ] common term outside

[ 10^3 + 1 ] * {(10^2 * C) + (10 * B) + A }

[ 10^3 + 1 ] = 1001 is divisible by 7. Hence A is sufficient.

Let f(1) = f(2) = ...... = f(6) = a ; then

B. (10^5 * a ) +(10^4 * a)+ ( 10^3 * a )+ (10^2 * a )+ (10 * a ) + a

taking a the common term outside ,

(10^5 + 10^4 + 10^3 + 10^2 + 10 + 1 ) a

111111 * a , here 111111 is divisible by 7. Hence B is sufficient. So, OA D.

I believe this helps.

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