Question

If X,Y and Z are positive numbers such that Y and Z have respectively 1 and 0 at their unit's place and $\Delta$ is the determinant $\left|\begin{array}{ccc}X& 4& 1\\ Y& 0& 1\\ Z& 1& 0\end{array}\right|$ If $(\Delta +1)$ is divisible by 10, then x has at its unit's place ?

• A. 1
• B. 0
• C. 2
• D. none of these

Answer 1

Answer: (C)

2

Let $X=10x+t$,$Y=10y+1$ and $Z=10z$ where $x,y,z$ are positive integers and $t$ is a non-negative integer.

Then $\Delta =\left|\begin{array}{ccc}10x+t& 4& 1\\ 10y+1& 0& 1\\ 10z& 1& 0\end{array}\right|=\left|\begin{array}{ccc}10x& 4& 1\\ 10y& 0& 1\\ 10z& 1& 0\end{array}\right|+\left|\begin{array}{ccc}t& 4& 1\\ 1& 0& 1\\ 0& 1& 0\end{array}\right|$
Therefore $\Delta =10k+(-t+1)$ where $k=\left|\begin{array}{ccc}x& 4& 1\\ y& 0& 1\\ z& 1& 0\end{array}\right|$
or $\Delta +1=10k+(2-t)$
Since $(\Delta +1)$ is divisible by $10$, Therefore $t=2$