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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 $|\begin{array}{lll} A 3 6 \\ 8 9 C \\ 2 B 2 \end{array}|$ is divisible by

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a

K²

b

K(K+1)

c

K+2

d

K

answer is C.

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Detailed Solution

$\begin{array}{l} |\begin{array}{ccc} A & 3 & 6 \\ 8 & 9 & 3 \\ 2 & B & 2 \end{array}| \\ R_{1} \to 100 R_{1} + 10 R_{3} + R_{2} \\ = |\begin{array}{ccc} A 28 & 3 B 9 & 62 C \\ 8 & 9 & 3 \\ 2 & B & 2 \end{array}| \\ = |\begin{array}{ccc} kx & ky & kz \\ 8 & 9 & 3 \\ 2 & B & 2 \end{array}| \end{array}$
Det is divisible by k

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