Let pqr be a three digit number. Then, pqr + qrp + rpq will not be always divisible by
9
pqr = 100p + 10q + r
qrp = 100q + 10r + p
rpq = 100r + 10p + q
Adding them we get:
100(p + q +r) + 10(p + q + r) + (p + q + r)
= 111 (p + q +r)
= 3 × 37 × (p + q + r)
Hence, (pqr + qrp + rpq) is divisible by 3, 37 and (p + q + r).
But it is not be divisible by 9 unless (p + q + r) is divisible by 9.