For any sequence of real number $A= \{ a_1, a_2, a_3,\dots\}$ for any sequence of real number $A= \{ a_1, a_2, a_3,\dots\}$ a sequence $\Delta A$ is defined as $\Delta A=\{a_2 - a_1, a_3 - a_2,\dots\}$ suppose that all terms of sequence $\Delta(\Delta A)$ are $1$ and that $a_{19}=a_{92}=0$ then find the digit at unit place of $a_3$ .
    I got the following equations but can't understand how to proceed further.
$$a_{21} - 2 a_{20}= 1$$
$$a_{17} - 2 a_{18}= 1$$
$$a_{18} + a_{20}=  1$$

Using a little theory, if the second difference is constant, the sequence is quadratic. If the constant is 1, the coefficient of the $n^2$-term is $1/2$. If $a_{19}=a_{92}=0$, then the quadratic has the factors $n-19$ and $n-92$. Now just put it all together. 
