Question

Consider a determinant det
$\left(B\right)=\left|\begin{array}{cc}(1003{)}^{12}& (1005{)}^{70}\\ (2016{)}^{212}& (2004{)}^{25}\end{array}\right|$ if |det(B)| is divided by 5, then remainder is

Answer 1

By binomial expansion
$(1003{)}^{12}=(1005-2{)}^{12}{=}^{12}{C}_0(1005{)}^{12}+...{+}^{12}{C}_{12}(-2{)}^{12}$
$=5k+2^{12}$ for some constant k
$=5k+4^6=5k+(5-1{)}^6$
=5(m)+1
$(1005{)}^{70}=5m,$
$(2016{)}^{212}=(2015+1{)}^{212}=5m_2+1$
$(2004{)}^{25}=(2005-1{)}^{25}=5m_3-1$
$det\left(B\right)=\left|\begin{array}{cc}5m+1& 5m_1\\ 5m_2+1& 5m_3-1\end{array}\right|$
$=\left|\begin{array}{cc}1& 0\\ 1& -1\end{array}\right|$ (after division by 5)
=(-1-0)
|det B| =1 so remainder is (1)