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 $100<p<1000$. The difference between maximum and minimum values of $(q+r)$, is	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 $N=(2+1)\left(2^2+1\right)\left(2^4+1\right)\dots \dots \left(2^{32}+1\right)+1$ 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)$\text{ Maximum if }p=3^0\cdot 2^9\Rightarrow (q+r{)}_{\max .}=9\left(2^9\to 512<1000\right)$
$\text{ Minimum if }p=3^3\cdot 2^2\Rightarrow (q+r{)}_{\min .}=5\left(3^5\cdot 2^0\to 243>100\right)$
$\text{ difference }=4$
(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 ${\log }_{8}a\ne 0$, 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 $2^{64}$
$\therefore {\log }_{2^8}{2}^{64}=8$