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"

m (→‎Solution 2)
m
Line 85: Line 85:
  
 
~mathboy100
 
~mathboy100
 +
 +
==See also==
 +
{{AIME box|year=2023|num-b=6|num-a=8|n=I}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 16:16, 8 February 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

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

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 $16 + 17 + 16 = \boxed{\textbf{(049) }}$.

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

Solution 2

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

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

Then, $n \equiv 2 \pmod{4}$, which implies $n \equiv 1 \pmod{3}$. By CRT, $n \equiv 4 \pmod{6}$, and therefore $n \equiv 3 \pmod{5}$. Using CRT again, 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 $n \equiv 0 \pmod{3}$, then by CRT, $n \equiv 3 \pmod{6}$, a contradiction. Thus, $n \equiv 2 \pmod{3}$, which by CRT implies $n \equiv 5 \pmod{6}$. $n$ can either be $0 \pmod{5}$, which implies that $n \equiv 35 \pmod{60}$, $17$ cases; or $4 \pmod{5}$, which implies that $n \equiv 59 \pmod{60}$, $16$ cases.

$16 + 16 + 17 = \boxed{49}$.

~mathboy100

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=189071"