Answer:
D. 9
Explanation:
Let
abc=100a+10b+c
bca=100b+10c+a
cab=100c+10a+b
abc + bca + cab=(100a+10b+c) + (100b+10c+a) + (100c+10a+b)
=100a + 10b + c + 100b + 10c + a + 100c + 10a + b
Collect like terms
=100a + a + 10a + 10b + 100b + b + c + 10c + 100c
=111a + 111b + 111 c
Factorise
=111(a+b+c)
abc + bca + cab = 111(a+b+c)
Factors of 111(a+b+c)= 1, 3, 37, 111, and (a+b+c)
abc + bca + cab is divisible by a+b+c because it is a factor of 111(a+b+c)
abc + bca + cab is divisible by 3 because 3 is a factor of 111(a+b+c)
abc + bca + cab is divisible by 37 because 37 is a factor of 111(a+b+c)
abc + bca + cab is not divisible by 9 because 9 is not a factor of 111(a+b+c)