Question

Let pqr be a three digit number. Then, pqr + qrp + rpq is always divisible by _____.

• A. 7
• B. 9
• C. 10
• D. 37

Answer 1

The correct option is D 37

pqr = 100p + 10q + r

qrp = 100q + 10r + p

rpq = 100r + 10p + q

Adding all three,
$\Rightarrow$ pqr + qrp + rpq
= 100(p + q +r) + 10(p + q + r) + (p + q + r) = 111 (p + q +r)
= 37 $\times$ 3 (p + q + r)

So, (pqr + qrp + rpq) is always divisible by 37, 3 and (p + q + r).