If X, Y and Z are positive numbers such that Y and Z have respectively 1 and 0 at their unit's place and Δ is the determinant $\begin{vmatrix}X &4& 1\\ Y&0&1\\Z& 1& 0\end{vmatrix}$

If (Δ + 1) is divisible by 10, then x has at its unit's place

2

Let $X = 10x+λ, Y = 10y+1$ and $Z = 10z$, where $x, y, z∈ N$. Then,

$Δ=\begin{vmatrix}X &4& 1\\ Y&0&1\\Z& 1& 0\end{vmatrix}$

$⇒Δ=\begin{vmatrix}10x+λ &4& 1\\ 10y+1&0&1\\10z+0& 1& 0\end{vmatrix}=\begin{vmatrix}10x &4& 1\\ 10y&0&1\\10z& 1& 0\end{vmatrix}+\begin{vmatrix}λ &4& 1\\ 1&0&1\\0& 1& 0\end{vmatrix}$

$⇒Δ=10\begin{vmatrix}x &4& 1\\ y&0&1\\z& 1& 0\end{vmatrix}+(1-λ)$

$⇒Δ+1=10\begin{vmatrix}x &4& 1\\ y&0&1\\z& 1& 0\end{vmatrix}+(2-λ)$

$⇒Δ+1=10k+(2-λ)$, where $k=\begin{vmatrix}x &4& 1\\ y&0&1\\z& 1& 0\end{vmatrix}$

It is given that Δ + 1 is divisible by 10. Therefore,

$2-λ=0$ i.e. $λ = 2$.

$∴X=10x + 2$

⇒ 2 is at unit's place of X.