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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 25"
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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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==Solution 3==
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To get from <math>n</math> to <math>n+1</math>, <math>R(n)</math> would add by <math>9</math> for each remainder <math>2, 3, 4, 5, 6, 7, 8, 9, 10</math>. However, given that some of these remainders can "round down" to <math>0</math> given the nature of mods, we must calculate the possible values of <math>n</math> such that the remainders in <math>R(n+1)</math> "rounds down" by a total of <math>9</math>, effectively canceling out the adding by <math>9</math> initially.
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To do so, we will analyze the "rounding down" for each of <math>2, 3, 4, 5, 6, 7, 8, 9, 10</math>:
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<math>n+1 \equiv 0 \pmod 2</math>:  subtract by <math>2</math>
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<math>n+1 \equiv 0 \pmod 3</math>:  subtract by <math>3</math>
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<math>n+1 \equiv 0 \pmod 4</math>:  subtract by <math>4</math>, but this also implies mod <math>2</math>, so subtract by <math>6</math>.
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<math>n+1 \equiv 0 \pmod 5</math>:  subtract by <math>5</math>
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<math>n+1 \equiv 0 \pmod 6</math>:  subtract by <math>6</math>, but this also implies mod <math>2</math> and <math>3</math>, so subtract by <math>11</math>: too much
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<math>n+1 \equiv 0 \pmod 7</math>:  subtract by <math>7</math>
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<math>n+1 \equiv 0 \pmod 8</math>:  subtract by <math>8</math>, but this also implies mod <math>2</math> and <math>4</math>, so subtract by <math>14</math>: too much
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<math>n+1 \equiv 0 \pmod 9</math>:  subtract by <math>9</math>, but this also implies mod <math>3</math>, so subtract by <math>12</math>: too much
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<math>n+1 \equiv 0 \pmod 10</math>:  subtract by <math>10</math>: too much
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Notice that <math>9 = 7+2 = 6+3 = 5+4 = 4+3+2</math>. By testing these sums, we can easily show that the only time when the total subtraction is <math>9</math> is when <math>n+1 \equiv 0 \pmod 2</math> AND <math>n+1 \equiv 0 \pmod 7</math>. By CRT, <math>n+1 \equiv 0 \pmod 14</math>:
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As in solution 1, then, 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>.
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~xHypotenuse
  
  

Revision as of 00:13, 21 August 2024

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 divisible by $2$, and $9$ itself indicates divisibility by $3$, too. So, $14|n+1$ and $n+1$ is not divisible 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)


Solution 3

To get from $n$ to $n+1$, $R(n)$ would add by $9$ for each remainder $2, 3, 4, 5, 6, 7, 8, 9, 10$. However, given that some of these remainders can "round down" to $0$ given the nature of mods, we must calculate the possible values of $n$ such that the remainders in $R(n+1)$ "rounds down" by a total of $9$, effectively canceling out the adding by $9$ initially.


To do so, we will analyze the "rounding down" for each of $2, 3, 4, 5, 6, 7, 8, 9, 10$:


$n+1 \equiv 0 \pmod 2$: subtract by $2$ $n+1 \equiv 0 \pmod 3$: subtract by $3$ $n+1 \equiv 0 \pmod 4$: subtract by $4$, but this also implies mod $2$, so subtract by $6$. $n+1 \equiv 0 \pmod 5$: subtract by $5$ $n+1 \equiv 0 \pmod 6$: subtract by $6$, but this also implies mod $2$ and $3$, so subtract by $11$: too much $n+1 \equiv 0 \pmod 7$: subtract by $7$ $n+1 \equiv 0 \pmod 8$: subtract by $8$, but this also implies mod $2$ and $4$, so subtract by $14$: too much $n+1 \equiv 0 \pmod 9$: subtract by $9$, but this also implies mod $3$, so subtract by $12$: too much $n+1 \equiv 0 \pmod 10$: subtract by $10$: too much


Notice that $9 = 7+2 = 6+3 = 5+4 = 4+3+2$. By testing these sums, we can easily show that the only time when the total subtraction is $9$ is when $n+1 \equiv 0 \pmod 2$ AND $n+1 \equiv 0 \pmod 7$. By CRT, $n+1 \equiv 0 \pmod 14$:


As in solution 1, then, only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions, so our answer is $\boxed{\textbf{(C) }2}$.


~xHypotenuse


Video Solution

https://youtu.be/vRKB4JdUIJ4

~MathProblemSolvingSkills.com


Video Solution by Mathematical Dexterity

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

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