JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 13)
If {p} denotes the fractional part of the number p, then
$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$, is equal to :
$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$, is equal to :
$${5 \over 8}$$
$${7 \over 8}$$
$${1 \over 8}$$
$${3 \over 8}$$
Explanation
$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$
= $$\left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{1 + 8K} \over 8}} \right\}$$
= $$\left\{ {{1 \over 8} + K} \right\}$$ where K $$ \in $$ Integer
$$ \therefore $$ Fractional part = $${{1 \over 8}}$$
= $$\left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \right\}$$
= $$\left\{ {{{1 + 8K} \over 8}} \right\}$$
= $$\left\{ {{1 \over 8} + K} \right\}$$ where K $$ \in $$ Integer
$$ \therefore $$ Fractional part = $${{1 \over 8}}$$
Comments (0)
