Question

Suppose a 3-digit number abc is divisible by 3. Prove that
abc+bca+cab is divisible by 9.​

Answer 1

Answer:

234

Step-by-step explanation:

234+342+423 =9+9+9=27 that is 27 is divisible by 9 and 3

Answer 2

Answer:

The proof is given below.

Step-by-step explanation:

Let the number abc is divided by 3.

Therefore, as per the divisibility rules of 3, a+b+c = 3*k, where k is an integer. --- Eq(1).

Now,from Eq(1), rearranging the LHS, we get,  b+c+a = 3*k. i.e. bca is divisible by 3.

Similarly, c+a+b = 3*k. i.e. cab is dvisible by 3.

Hence, abc+bca+cab = (100*a+10*b+c)+(100*b+10*c+a)+(100*c+10*a+b)

                                   = 100*(a+b+c)+10*(b+c+a)+(c+a+b)

                                   = 100*3*k +10*3*k+3*k

                                   = 333*k = 9*37*k, which is divisible by 9.

Hence, proved.

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