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Difference between revisions of "2023 AIME I Problems/Problem 7"

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==Solution 2 (Simpler)==
 
==Solution 2 (Simpler)==
  
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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.
+
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.
  
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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 (35).  
+
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>).  
  
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48 + 1 =  <math>\boxed{49}</math>.
+
<math>48 + 1 =  \boxed{49}</math>.
  
 
~Algebraik
 
~Algebraik

Revision as of 12:25, 9 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 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 $n \equiv 1 \pmod{3}$. 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

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 ($35$).

$48 + 1 = \boxed{49}$.

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

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

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