JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 14)
If the fractional part of the number $$\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$$, then k is equal to :
8
14
6
1
Explanation
$${{{2^{403}}} \over {15}}$$
$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$
$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$
$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$
$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$
$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$
$$ = {8 \over {15}} + 8$$ (integer)
$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$
According to the question,
$${k \over {15}} = {8 \over {15}}$$
$$ \Rightarrow $$ K $$=$$ 8
$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$
$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$
$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$
$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$
$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$
$$ = {8 \over {15}} + 8$$ (integer)
$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$
According to the question,
$${k \over {15}} = {8 \over {15}}$$
$$ \Rightarrow $$ K $$=$$ 8
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