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Difference between revisions of "2023 AIME I Problems/Problem 7"
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<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 = 0 \pmod{2}</math>
+
Case 1: <math>n \equiv 0 \pmod{2}</math>
  
−
Then, <math>n = 2 \pmod{4}</math>, which implies <math>n = 1 \pmod{3}</math>. By CRT, <math>n = 4 \pmod{6}</math>, and therefore <math>n = 3 \pmod{5}</math>. Using CRT again, we obtain <math>n = 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>. 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>.
  
−
Case 2: <math>n = 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</math> is <math>0 \pmod{3}</math>, then by CRT, <math>n = 3 \pmod{6}</math>, a contradiction. Thus, <math>n = 2 \pmod{3}</math>, which by CRT implies <math>n = 5 \pmod{6}</math>. <math>n</math> can either be <math>0 \pmod{5}</math>, which implies that <math>n = 35 \pmod{60}</math>, <math>17</math> cases; or <math>4 \pmod{5}</math>, which implies that <math>n = 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>, 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>16 + 16 + 17 = \boxed{49}</math>.
 
<math>16 + 16 + 17 = \boxed{49}</math>.
  
 
~mathboy100
 
~mathboy100

Revision as of 17:12, 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

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