Question

Check whether (abc + bca + cab) is always divisible by 37 or not?

(3 marks)

Answer 1

The generalised form of abc = 100×a + 10×b + 1×c
(0.5 marks)

Similarly, bca = 100×b + 10×c + 1×a
cab = 100×c + 10×a + 1×b.
(0.5 marks)

Adding these three numbers, we get
abc + bca + cab = (100×a + 10×b + 1×c) + (100×b + 10×c + 1×a) + (100×c + 10×a + 1×b).
⇒ (100+10+1)×a + (100+10+1)×b + (100+10+1)×c
⇒ abc + bca + cab = 111×a + 111×b + 111×c
⇒ abc + bca + cab = 111×(a+b+c).
(1 mark)

The number 111 can be factorised into 3 × 37.
(0.5 marks)

Therefore, abc + bca + cab = 37 × 3(a+b+c).
(0.5 marks)

So, we can say that (abc + bca + cab) is always divisble by 37