Question

The result obtained by addiing all three clockwise arrangements of three digits p, q and r, will always be divisible by ___.

Answer 1

Clockwise arrangments of 3 digits are

pqr = 100p + 10q + r

qrp = 100q + 10r + p

rpq = 100r + 10p + q

On adding, we get

100p + 10q + r + 100q + 10r + p + 100r + 10p + q

= 111p + 111q + 111r

= 111(p + q + r)

It has a multiple of 111. Hence, it is divisible by 111.