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Q.

If a1b1c1,a2b2c2 and a3b3c3 are three-digit numbers, each of which is divisible by a non zero integer k, then =a1b1c1a2b2c2a3b3c3 is always  
 

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a

divisible by k

b

divisible by k2

c

divisible by 2k

d

not divisible by k

answer is A.

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Detailed Solution

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Since a1b1c1,a2b2c2 and a3b3c3 are divisible by k, therefore

100a1+10b1+c1=n1k100a2+10b2+c2=n2k100a3+10b3+c3=n3k

where n1, n2, n3 are integers.

Now Δ=a1b1c1a2b2c2a3b3c3=a1b1100a1+10b1+c1a2b2100a2+10b2+c2a3b3100a3+10b3+c3

(Applying C3C3+10C2+100C1)

=a1b1n1ka2b2n2ka3b3n3k=Ka1b1n1a2b2n2a3b3n3=kΔ1

  Δ is divisible by k

(Since elements of Δ1 are integers,   Δ1 is an integer)

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