Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
(→‎Solution 4)
Line 159: Line 159:
  
 
==Solution 4==
 
==Solution 4==
−
Upon adding one to <math>n</math>, consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from <math>5\rightarrow 0</math> mod 6 and in general, <math>n-1 \rightarrow 0</math> mod n. We will use '<math>+1</math>' to refer to remainders that increase by 1, and 'wrap-arounds' to refer to remainders that go to 0.
+
Upon adding one to <math>n</math>, consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from <math>5\rightarrow 0</math> mod 6 and in general, <math>n-1 \rightarrow 0</math> mod n. We will use '<math>+1</math>' to refer to remainders that increase by 1, and 'wrap-around's to refer to remainders that go to 0.
  
  
Line 165: Line 165:
  
  
−
If there are <math>8</math> <math>+1</math>s and <math>1</math> wrap-arounds, the wrap-around must be equal to <math>-8</math>, which is the case for <math>(9)</math>. However, if <math>n</math> is <math>8</math> mod <math>9</math>, it clearly must also be <math>2</math> mod <math>3</math>, meaning there must more be one wrap-around, and this case won't work.
+
If there are <math>8</math> <math>+1</math>s and <math>1</math> wrap-around, the wrap-around must be equal to <math>-8</math>, which is the case for <math>(9)</math>. However, if <math>n</math> is <math>8</math> mod <math>9</math>, it clearly must also be <math>2</math> mod <math>3</math>, meaning there must more be one wrap-around, and this case won't work.
  
  

Revision as of 19:31, 4 November 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


Solution 4

Upon adding one to $n$, consider each individual remainder. Either it will increase by 1, or it will "wrap around', going from $5\rightarrow 0$ mod 6 and in general, $n-1 \rightarrow 0$ mod n. We will use '$+1$' to refer to remainders that increase by 1, and 'wrap-around's to refer to remainders that go to 0.


Clearly, $9$ $+1$s isn't possible, since then $R(n)\ne R(n+1)$.


If there are $8$ $+1$s and $1$ wrap-around, the wrap-around must be equal to $-8$, which is the case for $(9)$. However, if $n$ is $8$ mod $9$, it clearly must also be $2$ mod $3$, meaning there must more be one wrap-around, and this case won't work.


If there are $7$ $+1$s and $2$ wrap-arounds, these two wrap-arounds must add to $-7$. For the possible modulo, we could have $(2,7)$, $(3,6)$, and $(4,5)$. Clearly, $(3,6)$ won't work since if it is $5$ mod $6$, then it must also be $1$ mod $2$, meaning $(3,6)$ won't be the only wrap-arounds. Similarly, $(4,5)$ doesn't work since $3$ mod $4$ implies that $1$ mod $2$ will also be a wrap-around. That leaves $(2,7)$. The number must be $1$ mod $2$ and $6$ mod $7$, or in other words, $-1$ mod $2$ and $-1$ mod $7$, meaning n will be $-1 \equiv 13$ mod $14$. Testing all such two digits numbers that are equivalent to $13$ mod $14$, we see that $13$ and $97$ are the only two that work.


If there are $6$ $+1$s and $3$ wrap-arounds, the only possible combination of modulo is $(2,3,4)$. Thus, $n$ must be $11$ mod $12$. However, this means that mod $6$ will also be a wrap around, so this case won't work.


Notice that there can be no more cases, as for $5$ $+1$s, no matter what mods wrap around, the $+1$s will not be able to balance them out, as it's magnitude is too small. Therefore, there are only $\boxed{\textbf{C) }2}$ numbers.

Video Solution

https://youtu.be/vRKB4JdUIJ4

~MathProblemSolvingSkills.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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_25&oldid=231590"