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 "2023 AIME I Problems/Problem 7"

(→‎Solution 3)
(→‎Solution 4 (Small addition to solution 2))
 
(21 intermediate revisions by 7 users not shown)
Line 4: Line 4:
 
==Solution 1==
 
==Solution 1==
  
−
<math>n</math> can either be <math>0</math> or <math>1</math> mod <math>2</math>.
+
<math>n</math> can either be <math>0</math> or <math>1</math> (mod <math>2</math>).
  
 
Case 1: <math>n \equiv 0 \pmod{2}</math>
 
Case 1: <math>n \equiv 0 \pmod{2}</math>
  
−
Then, <math>n \equiv 2 \pmod{4}</math>, which implies <math>n \equiv 1 \pmod{3}</math>. By CRT, <math>n \equiv 4 \pmod{6}</math>, and therefore <math>n \equiv 3 \pmod{5}</math>. Using CRT again, we obtain <math>n \equiv 58 \pmod{60}</math>, which gives <math>16</math> values for <math>n</math>.
+
Then, <math>n \equiv 2 \pmod{4}</math>, which implies <math>n \equiv 1 \pmod{3}</math> and <math>n \equiv 4 \pmod{6}</math>, and therefore <math>n \equiv 3 \pmod{5}</math>. Using [[Chinese Remainder Theorem|CRT]], we obtain <math>n \equiv 58 \pmod{60}</math>, which gives <math>16</math> values for <math>n</math>.
  
 
Case 2: <math>n \equiv 1 \pmod{2}</math>
 
Case 2: <math>n \equiv 1 \pmod{2}</math>
  
−
<math>n</math> is then <math>3 \pmod{4}</math>. If <math>n \equiv 0 \pmod{3}</math>, then by CRT, <math>n \equiv 3 \pmod{6}</math>, a contradiction. Thus, <math>n \equiv 2 \pmod{3}</math>, which by CRT implies <math>n \equiv 5 \pmod{6}</math>. <math>n</math> can either be <math>0 \pmod{5}</math>, which implies that <math>n \equiv 35 \pmod{60}</math>, <math>17</math> cases; or <math>4 \pmod{5}</math>, which implies that <math>n \equiv 59 \pmod{60}</math>, <math>16</math> cases.
+
<math>n</math> is then <math>3 \pmod{4}</math>. If <math>n \equiv 0 \pmod{3}</math>, <math>n \equiv 3 \pmod{6}</math>, a contradiction. Thus, <math>n \equiv 2 \pmod{3}</math>, which implies <math>n \equiv 5 \pmod{6}</math>. <math>n</math> can either be <math>0 \pmod{5}</math>, which implies that <math>n \equiv 35 \pmod{60}</math> by CRT, giving <math>17</math> cases; or <math>4 \pmod{5}</math>, which implies that <math>n \equiv 59 \pmod{60}</math> by CRT, giving <math>16</math> cases.
  
−
<math>16 + 16 + 17 = \boxed{49}</math>.
+
The total number of extra-distinct numbers is thus <math>16 + 16 + 17 = \boxed{049}</math>.
  
 
~mathboy100
 
~mathboy100
  
−
==Solution 2==
+
==Solution 2 (Simpler)==
 +
 
 +
Because the LCM of all of the numbers we are dividing by is <math>60</math>, we know that all of the remainders are <math>0</math> again at <math>60</math>, meaning that we have a cycle that repeats itself every <math>60</math> numbers.
 +
 
 +
After listing all of the remainders up to <math>60</math>, we find that <math>35</math>, <math>58</math>, and <math>59</math> are extra-distinct. So, we have <math>3</math> numbers every <math>60</math> which are extra-distinct. <math>60\cdot16 = 960</math> and <math>3\cdot16 = 48</math>, so we have <math>48</math> extra-distinct numbers in the first <math>960</math> numbers. Because of our pattern, we know that the numbers from <math>961</math> thru <math>1000</math> will have the same remainders as <math>1</math> thru <math>40</math>, so we have <math>1</math> other extra-distinct number (<math>35</math>).
 +
 
 +
<math>48 + 1 =  \boxed{049}</math>.
 +
 
 +
~Algebraik
 +
 
 +
==Solution 3==
  
 
<math>\textbf{Case 0:  } {\rm Rem} \ \left( n, 6 \right) = 0</math>.
 
<math>\textbf{Case 0:  } {\rm Rem} \ \left( n, 6 \right) = 0</math>.
Line 81: Line 91:
 
Therefore, the number extra-distinct <math>n</math> in this subcase is 16.
 
Therefore, the number extra-distinct <math>n</math> in this subcase is 16.
  
−
Putting all cases together, the total number of extra-distinct <math>n</math> is <math>16 + 17 + 16 = \boxed{\textbf{(049) }}</math>.
+
Putting all cases together, the total number of extra-distinct <math>n</math> is <math>16 + 17 + 16 = \boxed{049}</math>.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 +
==Solution 4 (Small addition to solution 2)==
 +
We need to find that <math>35</math>, <math>58</math>, and <math>59</math> are all extra-distinct numbers smaller then <math>61.</math>
 +
 +
Let <math>k \in \left\{2, 3, 4, 5, 6 \right\}.</math> Denote the remainder in the division of <math>a</math> by <math>b</math> as <math>{\rm Rem} \ \left( a, b \right).</math>
 +
 +
<math>{\rm Rem} \ \left( -1, k \right) = k - 1 \implies {\rm Rem} \ \left( 59, k \right) = k - 1 = \left\{1, 2, 3, 4, 5 \right\}\implies 59</math> is extra-distinct.
 +
<math>{\rm Rem} \ \left( -2, k \right) = k - 2 \implies {\rm Rem} \ \left( 58, k \right) = k - 2 = \left\{0, 1, 2, 3, 4 \right\} \implies 58</math> is extra-distinct.
 +
<cmath>{\rm Rem} \ \left( x + 12y, k \right) = {\rm Rem} \ \left(x, k \right) + \left\{0, 0, 0, {\rm Rem} \ \left(12y, k \right), 0 \right\}.</cmath>
 +
We need to check all of the remainders up to <math>12 - 3 = 9</math> and remainders
 +
<cmath>{\rm Rem} \ \left( 59 - 12, k \right) = {\rm Rem} \ \left( 59 - 36, k \right) = \left\{1, 2, 3, 3, 5 \right\},
 +
{\rm Rem} \ \left( 59 - 48, k \right) = \left\{1, 2, 3, 1, 5 \right\},</cmath>
 +
<math>{\rm Rem} \ \left( 59 - 24, k \right) ={\rm Rem} \ \left(35, k \right) = \left\{1, 2, 3, 0, 5 \right\} \implies 35</math> is extra-distinct.
 +
<math>58 - 12 = 46 \implies {\rm Rem} \ \left( 46, 5 \right) = 1 = {\rm Rem} \ \left( 46, 3 \right), </math>
 +
<math>58 - 24 = 34 \implies {\rm Rem} \ \left( 34, 5 \right) = 4 = {\rm Rem} \ \left( 34, 6 \right), </math>
 +
<math>58 - 36 = 22 \implies {\rm Rem} \ \left( 22, 5 \right) = 2 = {\rm Rem} \ \left( 22, 4 \right), </math>
 +
<math>58 - 48 = 10 \implies {\rm Rem} \ \left( 10, 5 \right) = 0 = {\rm Rem} \ \left( 10, 2 \right). </math>
  
−
==Solution 3==
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
−
 
 
−
Because the LCM of all of the numbers we are dividing by is 60, we know that all of the remainders are 0 again at 60, meaning that we have a cycle that repeats itself every 60 numbers.
 
  
−
After listing all of the remainders up to 60, we find that 35, 58, and 59 are extra-distinct. So, we have 3 numbers every 60 which are extra-distinct. <math>60\cdot16</math> = 960 and <math>3\cdot16</math> = 48, so we have 48 extra-distinct numbers in the first 960 numbers. Because of our pattern, we know that the numbers from 961-1000 will act the same as 1-40, where we have 1 extra-distinct number. 48 + 1 = <math>\boxed{49}</math>.
+
==Video Solution==
 +
https://youtu.be/8oOik9d1fWM
  
−
-Algebraik
+
~MathProblemSolvingSkills.com
  
 
==See also==
 
==See also==

Latest revision as of 16:55, 31 October 2023

Problem

Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$.

Solution 1

$n$ can either be $0$ or $1$ (mod $2$).

Case 1: $n \equiv 0 \pmod{2}$

Then, $n \equiv 2 \pmod{4}$, which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$, and therefore $n \equiv 3 \pmod{5}$. Using CRT, we obtain $n \equiv 58 \pmod{60}$, which gives $16$ values for $n$.

Case 2: $n \equiv 1 \pmod{2}$

$n$ is then $3 \pmod{4}$. If $n \equiv 0 \pmod{3}$, $n \equiv 3 \pmod{6}$, a contradiction. Thus, $n \equiv 2 \pmod{3}$, which implies $n \equiv 5 \pmod{6}$. $n$ can either be $0 \pmod{5}$, which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$, which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.

The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$.

~mathboy100

Solution 2 (Simpler)

Because the LCM of all of the numbers we are dividing by is $60$, we know that all of the remainders are $0$ again at $60$, meaning that we have a cycle that repeats itself every $60$ numbers.

After listing all of the remainders up to $60$, we find that $35$, $58$, and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\cdot16 = 960$ and $3\cdot16 = 48$, so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$, so we have $1$ other extra-distinct number ($35$).

$48 + 1 = \boxed{049}$.

~Algebraik

Solution 3

$\textbf{Case 0: } {\rm Rem} \ \left( n, 6 \right) = 0$.

We have ${\rm Rem} \ \left( n, 2 \right) = 0$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution.

$\textbf{Case 1: } {\rm Rem} \ \left( n, 6 \right) = 1$.

We have ${\rm Rem} \ \left( n, 2 \right) = 1$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution.

$\textbf{Case 2: } {\rm Rem} \ \left( n, 6 \right) = 2$.

We have ${\rm Rem} \ \left( n, 3 \right) = 2$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution.

$\textbf{Case 3: } {\rm Rem} \ \left( n, 6 \right) = 3$.

The condition ${\rm Rem} \ \left( n, 6 \right) = 3$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$, ${\rm Rem} \ \left( n, 3 \right) = 0$.

Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 \right\}$ is a permutation of $\left\{ 0, 1 ,2 \right\}$. Thus, ${\rm Rem} \ \left( n, 4 \right) = 2$.

However, ${\rm Rem} \ \left( n, 4 \right) = 2$ conflicts ${\rm Rem} \ \left( n, 2 \right) = 1$. Therefore, this case has no solution.

$\textbf{Case 4: } {\rm Rem} \ \left( n, 6 \right) = 4$.

The condition ${\rm Rem} \ \left( n, 6 \right) = 4$ implies ${\rm Rem} \ \left( n, 2 \right) = 0$ and ${\rm Rem} \ \left( n, 3 \right) = 1$.

Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 , 5 \right\}$ is a permutation of $\left\{ 0, 1 ,2 , 3 \right\}$.

Because ${\rm Rem} \ \left( n, 2 \right) = 0$, we must have ${\rm Rem} \ \left( n, 4 \right) = 2$. Hence, ${\rm Rem} \ \left( n, 5 \right) = 3$.

Hence, $n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}$. Hence, $n \equiv - 2 \pmod{60}$.

We have $1000 = 60 \cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this case is 16.

$\textbf{Case 5: } {\rm Rem} \ \left( n, 6 \right) = 5$.

The condition ${\rm Rem} \ \left( n, 6 \right) = 5$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ and ${\rm Rem} \ \left( n, 3 \right) = 2$.

Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, 4 \right)$ and ${\rm Rem} \ \left( n, 5 \right)$ are two distinct numbers in $\left\{ 0, 3, 4 \right\}$. Because ${\rm Rem} \ \left( n, 4 \right) \leq 3$ and $n$ is odd, we have ${\rm Rem} \ \left( n, 4 \right) = 3$. Hence, ${\rm Rem} \ \left( n, 5 \right) = 0$ or 4.

$\textbf{Case 5.1: } {\rm Rem} \ \left( n, 6 \right) = 5$, ${\rm Rem} \ \left( n, 4 \right) = 3$, ${\rm Rem} \ \left( n, 5 \right) = 0$.

We have $n \equiv 35 \pmod{60}$.

We have $1000 = 60 \cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this subcase is 17.

$\textbf{Case 5.2: } {\rm Rem} \ \left( n, 6 \right) = 5$, ${\rm Rem} \ \left( n, 4 \right) = 3$, ${\rm Rem} \ \left( n, 5 \right) = 4$.

$n \equiv - 1 \pmod{60}$.

We have $1000 = 60 \cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this subcase is 16.

Putting all cases together, the total number of extra-distinct $n$ is $16 + 17 + 16 = \boxed{049}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 4 (Small addition to solution 2)

We need to find that $35$, $58$, and $59$ are all extra-distinct numbers smaller then $61.$

Let $k \in \left\{2, 3, 4, 5, 6 \right\}.$ Denote the remainder in the division of $a$ by $b$ as ${\rm Rem} \ \left( a, b \right).$

${\rm Rem} \ \left( -1, k \right) = k - 1 \implies {\rm Rem} \ \left( 59, k \right) = k - 1 = \left\{1, 2, 3, 4, 5 \right\}\implies 59$ is extra-distinct. ${\rm Rem} \ \left( -2, k \right) = k - 2 \implies {\rm Rem} \ \left( 58, k \right) = k - 2 = \left\{0, 1, 2, 3, 4 \right\} \implies 58$ is extra-distinct. \[{\rm Rem} \ \left( x + 12y, k \right) = {\rm Rem} \ \left(x, k \right) + \left\{0, 0, 0, {\rm Rem} \ \left(12y, k \right), 0 \right\}.\] We need to check all of the remainders up to $12 - 3 = 9$ and remainders \[{\rm Rem} \ \left( 59 - 12, k \right) = {\rm Rem} \ \left( 59 - 36, k \right) = \left\{1, 2, 3, 3, 5 \right\}, {\rm Rem} \ \left( 59 - 48, k \right) = \left\{1, 2, 3, 1, 5 \right\},\] ${\rm Rem} \ \left( 59 - 24, k \right) ={\rm Rem} \ \left(35, k \right) = \left\{1, 2, 3, 0, 5 \right\} \implies 35$ is extra-distinct. $58 - 12 = 46 \implies {\rm Rem} \ \left( 46, 5 \right) = 1 = {\rm Rem} \ \left( 46, 3 \right),$ $58 - 24 = 34 \implies {\rm Rem} \ \left( 34, 5 \right) = 4 = {\rm Rem} \ \left( 34, 6 \right),$ $58 - 36 = 22 \implies {\rm Rem} \ \left( 22, 5 \right) = 2 = {\rm Rem} \ \left( 22, 4 \right),$ $58 - 48 = 10 \implies {\rm Rem} \ \left( 10, 5 \right) = 0 = {\rm Rem} \ \left( 10, 2 \right).$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/8oOik9d1fWM

~MathProblemSolvingSkills.com

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2023_AIME_I_Problems/Problem_7&oldid=200550"