P is a two-digit integer, which can be written as 30a + b, where a and

P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.

(1) a =4
(2) b3–5b2–14b=0

Given : P=30a + b ; divide by 3
\(\frac{P}{3}\) = \(\frac{30a}{3}\) + \(\frac{b}{3}\)
Now, 30 is divisible by 3, hence 30a will leave a remainder 0 upon division by 3; so actual remainder will be given by value of \(\frac{b}{3}\)
Question now becomes b=?

Stat1) a=4 , gives nothing about b, hence Not suff

Stat2) \(b^3–5b^2–14b=0\), take common b out => \(b*(b^2-5b-14)=0\) so either b=0 or \(b^2-5b-14=0\) ; further solving the later we get b= 7,-2. Since it is given that b is positive so in all we get b is 7; So remainder will be 1 Hence Suff
B

Edited : Corrected my mistake of missing that b was positive so 0 wouldnt count anyway .. silly me