Question

Column I		Column II
A	When the repeating decimal 0.363636 . is written as a rational fraction in the simplest form, the sum of the numerator and denominator is	P	4
B	Given positive integer $p, q$ and $r$ with $p=3^q \cdot 2^r$ and $100	Q	8
C	If $\log _8 a+\log _8 b=\left(\log _8 a\right)\left(\log _8 b\right)$ and $\log _a b=3$, then the value of ' $a$ ' is	R	15
D	Let $N =(2+1)\left(2^2+1\right)\left(2^4+1\right) \ldots \ldots\left(2^{32}+1\right)+1$ then $\log _{256} N$ equals	S	16

• A. (A) R; (B) P; (C) S; (D) P
• B. (A) R; (B) P; (C) S; (D) Q
• C. (A) R; (B) P; (C) S; (D) R
• D. (A) R; (B) P; (C) S; (D) S

Answer 1

Answer: (B)

(A) $x =\frac{36}{99}=\frac{4}{11} \Rightarrow 15$ Ans.
(B)$\text { Maximum if } p =3^0 \cdot 2^9 \Rightarrow ( q + r )_{\max .}=9\left(2^9 \rightarrow 512<1000\right) $
$\text { Minimum if } p =3^3 \cdot 2^2 \Rightarrow ( q + r )_{\min .}=5\left(3^5 \cdot 2^0 \rightarrow 243>100\right)$
$\text { difference }=4 $
(C) $\log _a b=3 \Rightarrow \frac{\log _8 b}{\log _8 a}=3 \Rightarrow \log _8 b=3 \log _8 a$
Substituting in $\log _8 a+\log _8 b=\log _8 a \cdot \log _8 b$ gives $4 \log _8 a=3\left(\log _8 a\right)^2$. Since $\log _8 a \neq 0$, then $4=3 \log _8 a \Rightarrow \log _8 a =\frac{4}{3}$ so $a =\left(2^3\right)^{4 / 3}=16$ Ans. ]
(D) $N$ simplifies to $2^{64}$
$\therefore \log _{2^8} 2^{64}=8$