Question

The remainder when 31^150 is divided by 1024 is. (a)320
(b)704
(c)703
(d)321​

Answer 1

$\textbf{Given:}$

$\text{Number is 31^{150} }$

$\textbf{To find:}$

$\text{The remainder when 31^{150} is divided by 1024}$

$\textbf{Solution:}$

$\text{Consider,}$

$31^{150}$

$=(32-1)^{150}$

$\text{Expanding by using binomial theorem}$

$={150}_C_0\,(-1)^0(32)^{150}+{150}_C_1\,(-1)^1(32)^{149}+{150}_C_2\,(-1)^2(32)^{148}+................+{150}_C_{148}\,(-1)^{148}(32)^2+{150}_C_{149}\,(-1)^{149}(32)^1+{150}_C_{150}\,(-1)^{150}(32)^0$

$={150}_C_0\,(32)^{150}-{150}_C_1\,(32)^{149}+{150}_C_2\,(32)^{148}+................+{150}_C_{148}\,(32)^2-{150}_C_{149}\,(32)^1+{150}_C_{150}$

$\text{Using,}$

$\bf\,n_C_r=n_C_{n-r}$

$\bf\,n_C_1=n$

$\bf\,n_C_n=1$

$=[{150}_C_0\,(32)^{148}(32)^2-{150}_C_1\,(32)^{147}(32)^2+{150}_C_2\,(32)^{146}(32)^2+................+{150}_C_{148}\,(32)^2]-{150}_C_1\,(32)^1+1$

$=[{150}_C_0\,(32)^{148}(1024)-{150}_C_1\,(32)^{147}(1024)^+{150}_C_2\,(32)^{146}(1024)+................+{150}_C_{148}\,(1024)]-(150{\times}32)+1$

$=[\textbf{a positive multiple of 1024}]-4800+1$

$=[k{\times}1024]-4799$

$\text{It is clear that k{\times}1024 is a multiple of 1024 greater than 5120}$

$\text{We have}$

$=[l{\times}1024+5120]-4799$

$=l{\times}1024+(5120-4799)$

$=l{\times}1024+321$

$\implies\,31^{150}=l{\times}1024+321$

$\therefore\textbf{The remainder when 31^{150} is divided by 1024 is 321}$

$\textbf{Answer:}$

$\textbf{Option (d) is correct}$