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Question

If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -

  1. 6    
  2. 12    
  3. 24    
  4. None of these    

The correct answer is: 6


    We have (a3 – a) + (b3 – b) + (c3 – c)
    = (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
    Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
    6 | {(a3 – a) + (b3 – b) + (c3 – c)}
    rightwards double arrow6 | (a3 + b3 + c3) as 6|(a + b + c).

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