Difference between revisions of "2021 Fall AMC 12B Problems/Problem 25"

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Note that we can add <math>9</math> to <math>R(n)</math> to get <math>R(n+1)</math>, but must subtract <math>k</math> for all <math>k|n+1</math>. Hence, we see that there are four ways to do that because <math>9=7+2=6+3=5+4=4+3+2</math>. Note that only <math>7+2</math> is a plausible option, since <math>4+3+2</math> indicates <math>n+1</math> is divisible by <math>6</math>, <math>5+4</math> indicates that <math>n+1</math> is divisible by <math>2</math>, <math>6+3</math> indicates <math>n+1</math> is divisibly by <math>2</math>, and <math>9</math> itself indicates divisibility by <math>3</math>, too. So, <math>14|n+1</math> and <math>n+1</math> is not divisibly by any positive integers from <math>2</math> to <math>10</math>, inclusive, except <math>2</math> and <math>7</math>. We check and get that only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions so our answer is <math>\boxed{\textbf{(C) }2}</math>.
 
Note that we can add <math>9</math> to <math>R(n)</math> to get <math>R(n+1)</math>, but must subtract <math>k</math> for all <math>k|n+1</math>. Hence, we see that there are four ways to do that because <math>9=7+2=6+3=5+4=4+3+2</math>. Note that only <math>7+2</math> is a plausible option, since <math>4+3+2</math> indicates <math>n+1</math> is divisible by <math>6</math>, <math>5+4</math> indicates that <math>n+1</math> is divisible by <math>2</math>, <math>6+3</math> indicates <math>n+1</math> is divisibly by <math>2</math>, and <math>9</math> itself indicates divisibility by <math>3</math>, too. So, <math>14|n+1</math> and <math>n+1</math> is not divisibly by any positive integers from <math>2</math> to <math>10</math>, inclusive, except <math>2</math> and <math>7</math>. We check and get that only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions so our answer is <math>\boxed{\textbf{(C) }2}</math>.
  
-kevinmathz
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- kevinmathz
  
 
{{AMC12 box|year=2021 Fall|ab=B|num-b=24|after=Last problem}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-b=24|after=Last problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:08, 25 November 2021

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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All AMC 12 Problems and Solutions

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