If abc,lmn and pqr be any three-digit numbers each of which is divisible by k, then show that Δ=∣∣
∣∣abclmnpqr∣∣
∣∣ is also divisible by k.
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Solution
abc=100a+10b+c=n1k etc. Apply C3+10C2+100C1 Δ=∣∣
∣∣abn1klmn2kpqn3k∣∣
∣∣=k∥=kΔ1 where Δ1 is a determinant consisting of integers. ∴kΔ1 is also an integer. Hence Δ is divisible by k