2021 Fall AMC 12B Problems/Problem 25

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 1

Note that we can add $9$ to $R(n)$ to get $R(n+1)$ , but must subtract $k$ for all $k|n+1$ . Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$ . Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$ , $5+4$ indicates that $n+1$ is divisible by $2$ , $6+3$ indicates $n+1$ is divisibly by $2$ , and $9$ itself indicates divisibility by $3$ , too. So, $14|n+1$ and $n+1$ is not divisibly by any positive integers from $2$ to $10$ , inclusive, except $2$ and $7$ . We check and get that only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions so our answer is $\boxed{\textbf{(C) }2}$ .

- kevinmathz

Solution 2

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 $\boxed{\textbf{(C) }2}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=Fy8wU4VAzkQ

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 Fall AMC 12B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

Video Solution by Mathematical Dexterity