JEE MAIN - Mathematics (2009 - No. 17)
The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is :
2
7
8
0
Giải thích
$${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$
= $${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$$
= $${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$$
= $$\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$$
+ $$\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$$
= 2 + 63$$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$$
= 63$$ \times $$[Some integral value] + 2
63$$ \times $$[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.
= $${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$$
= $${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$$
= $$\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$$
+ $$\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$$
= 2 + 63$$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$$
= 63$$ \times $$[Some integral value] + 2
63$$ \times $$[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.