Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.
\[\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 2 & & 2 & & 2 & & & \\ & & 3 & & 4 & & 4 & & 3 & & \\ & 4 & & 7 & & 8 & & 7 & & 4 & \\ 5 & & 11 & & 15 & & 15 & & 11 & & 5 \end{tabular}\]
Let $f(n)$ denote the sum of the numbers in row $n$. What is the remainder when $f(100)$ is divided by 100?
$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 }$
Note that if we re-draw the table with an additional diagonal row on each side, the table is actually just two of [Pascal's Triangles](https://artofproblemsolving.com/wiki/index.php/Pascal%27s_Triangle), except translated and summed. 
\[\begin{tabular}{ccccccccccccccc} & & & & & 1 & & 0 & & 1 & & & & \\ & & & & 1 & & 1 & & 1 & & 1 & & & \\ & & & 1 & & 2 & & 2 & & 2 & & 1 & & \\ & & 1 & & 3 & & 4 & & 4 & & 3 & & 1 & \\ & 1 & & 4 & & 7 & & 8 & & 7 & & 4 & & 1  \\ 1 & & 5 & & 11 & & 15 & & 15 & & 11 & & 5 & & 1 \end{tabular}\]
The sum of a row of Pascal's triangle is $2^{n-1}$; the sum of two of each of these rows, subtracting away the $2$ ones we included, yields $f(n) = 2^n - 2$. Now, $f(100) = 2^{100} - 2 \equiv 2 \pmod{4}$ and $f(100) = 2^{100} - 2 \equiv 2^{20 \cdot 5} - 2 \equiv -1 \pmod{25}$, and by the [Chinese Remainder Theorem](https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem), we have $f(100) \equiv 74 \pmod{100} \Longrightarrow \mathrm{(E)}$.

Remark (Chinese Remainder Theorem)
Solution 1 needs to calculate the last $2$ digits of $2^{100}$, meaning to solve $2^{100} (\bmod{ 100})$. $100 = 4 \cdot 25$, we are going to get $n \bmod{ 100}$ from $n \bmod{ 4}$ and $n \bmod{ 25}$.
By Chinese Remainder Theorem, the general solution of the system of $2$ linear congruences is:

$n \equiv r_1 (\bmod { \quad m_1})$, $n \equiv r_2 (\bmod { \quad m_2})$, $(m_1, m_2) = 1$
Find $k_1$ and $k_2$ such that $k_1 m_1 \equiv 1 (\bmod{ \quad m_2})$, $k_2 m_2 \equiv 1 (\bmod{ \quad m_1})$
Then $n \equiv k_2 m_2 r_1 + k_1 m_1 r_2 (\bmod{ \quad m_1 m_2})$

In this problem, $2^{100} \equiv (2^{10})^{10} \equiv 1024^{10} \equiv 24^{10} \equiv (-1)^{10} \equiv 1 (\bmod{ \quad 25})$, $2^{100} \equiv 0 (\bmod{ \quad 4})$:

$n \equiv 1 (\bmod{ \quad 25})$, $n \equiv 0 (\bmod{ \quad 4})$, $(25, 4) = 1$
$1 \cdot 25 \equiv 1 (\bmod{ \quad 4})$, $19 \cdot 4 \equiv 1 (\bmod{ \quad 25})$
Then $n \equiv 19 \cdot 4 \cdot 1 + 1 \cdot 25 \cdot 0 \equiv 76 (\bmod{ \quad 100})$

~[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)