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

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

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

Problem

Solution