Problem: For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ . For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$ . How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: Denote by ${\rm Rem} \ \left( n, k \right)$ the remainder of $n$ divided by $k$ .
Define $\Delta \left( n, k \right) = {\rm Rem} \ \left( n + 1, k \right) - {\rm Rem} \ \left( n, k \right)$
Hence, \[ \Delta \left( n, k \right) = \left\{ \begin{array}{ll} 1 & \mbox{ if } n \not\equiv -1 \pmod{k} \\ - \left( k  -1 \right) & \mbox{ if } n \equiv -1 \pmod{k} \end{array} \right.. \]
Hence, this problem asks us to find all $n \in \left\{ 10 , 11, \cdots , 99 \right\}$ , such that $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$
$\textbf{Case 1}$ $\Delta \left( n, 10 \right) = - 9$
We have $\sum_{k = 2}^9 \Delta \left( n, k \right) \leq \sum_{k = 2}^9 1 = 8$
Therefore, there is no $n$ in this case.
$\textbf{Case 2}$ $\Delta \left( n, 10 \right) = 1$ and $\Delta \left( n, 9 \right) = -8$
The condition $\Delta \left( n, 9 \right) = -8$ implies $n \equiv - 1 \pmod{9}$ .
This further implies $n \equiv - 1 \pmod{3}$ .
Hence, $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}}  1 = 6$
Therefore, there is no $n$ in this case.
$\textbf{Case 3}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 9 , 10 \right\}$ and $\Delta \left( n, 8 \right) = -7$
The condition $\Delta \left( n, 8 \right) = -7$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 4 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 4 \right) = -3$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 3, 5, 6, 7 \right\}}  1 = 4$
Therefore, there is no $n$ in this case.
$\textbf{Case 4}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 8, \cdots , 10 \right\}$ and $\Delta \left( n, 7 \right) = -6$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4, 5, 6 \right\}} \Delta \left( n, k \right) = 3$
Hence, we must have $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 3 , 4 , 5 , 6 \right\}$
Therefore, $n = 13, 97$
$\textbf{Case 5}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 7 , \cdots , 10 \right\}$ and $\Delta \left( n, 6 \right) = -5$
The condition $\Delta \left( n, 6 \right) = -5$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 3 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) = 4$
However, we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 4, 5 \right\}}  1 = 2$
Therefore, there is no $n$ in this case.
$\textbf{Case 6}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 6 , \cdots , 10 \right\}$ and $\Delta \left( n, 5 \right) = -4$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4 \right\}} \Delta \left( n, k \right) = -1$
This can be achieved if $\Delta \left( n, 2 \right) = 1$ $\Delta \left( n, 3 \right) = 1$ $\Delta \left( n, 4 \right) = -3$
However, $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{4}$ . This implies $n \equiv -1 \pmod{2}$ . Hence, $\Delta \left( n, 2 \right) = -1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 7}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 5 , \cdots , 10 \right\}$ and $\Delta \left( n, 4 \right) = -3$
The condition $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{k}$ with $k = 2$ .
Hence, $\Delta \left( n, 2 \right) = -1$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 3 \right) = - 2$ . This implies $n \equiv - 1 \pmod{3}$
Because $n \equiv - 1 \pmod{2}$ and $n \equiv - 1 \pmod{3}$ , we have $n \equiv - 1 \pmod{6}$ .
Hence, $\Delta \left( n, 6 \right) = - 5$ .
However, in this case, we assume $\Delta \left( n, 6 \right) = 1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 8}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 4 , \cdots , 10 \right\}$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 5$ . This is infeasible.
Therefore, there is no $n$ in this case.
$\textbf{Case 9}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{3 , \cdots , 10 \right\}$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 8$ . This is infeasible.
Therefore, there is no $n$ in this case.
Putting all cases together, the answer is $\boxed{2}$

Answer: C 2