Difference between revisions of "2023 AIME I Problems/Problem 7"
m (→Solution) |
|||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | \textbf{Case 0:} | + | <math>\textbf{Case 0:} {\rm Rem} \ \left( n, 6 \right) = 0</math>. |
We have <math>{\rm Rem} \ \left( n, 2 \right) = 0</math>. This violates the condition that <math>n</math> is extra-distinct. | We have <math>{\rm Rem} \ \left( n, 2 \right) = 0</math>. This violates the condition that <math>n</math> is extra-distinct. | ||
Therefore, this case has no solution. | Therefore, this case has no solution. | ||
| − | \textbf{Case 1:} | + | <math>\textbf{Case 1:} {\rm Rem} \ \left( n, 6 \right) = 1</math>. |
We have <math>{\rm Rem} \ \left( n, 2 \right) = 1</math>. This violates the condition that <math>n</math> is extra-distinct. | We have <math>{\rm Rem} \ \left( n, 2 \right) = 1</math>. This violates the condition that <math>n</math> is extra-distinct. | ||
Therefore, this case has no solution. | Therefore, this case has no solution. | ||
| − | \textbf{Case 2:} | + | #\textbf{Case 2:} {\rm Rem} \ \left( n, 6 \right) = 2<math>. |
| − | We have <math>{\rm Rem} \ \left( n, 3 \right) = 2< | + | We have </math>{\rm Rem} \ \left( n, 3 \right) = 2<math>. This violates the condition that </math>n<math> is extra-distinct. |
Therefore, this case has no solution. | Therefore, this case has no solution. | ||
| − | \textbf{Case 3:} | + | </math>\textbf{Case 3:} {\rm Rem} \ \left( n, 6 \right) = 3<math>. |
| − | The condition <math>{\rm Rem} \ \left( n, 6 \right) = 3< | + | The condition </math>{\rm Rem} \ \left( n, 6 \right) = 3<math> implies </math>{\rm Rem} \ \left( n, 2 \right) = 1<math>, </math>{\rm Rem} \ \left( n, 3 \right) = 0<math>. |
| − | Because <math>n< | + | Because </math>n<math> is extra-distinct, </math>{\rm Rem} \ \left( n, l \right)<math> for </math>l \in \left\{ 2, 3, 4 \right\}<math> is a permutation of </math>\left\{ 0, 1 ,2 \right\}<math>. |
| − | Thus, <math>{\rm Rem} \ \left( n, 4 \right) = 2< | + | Thus, </math>{\rm Rem} \ \left( n, 4 \right) = 2<math>. |
| − | However, <math>{\rm Rem} \ \left( n, 4 \right) = 2< | + | However, </math>{\rm Rem} \ \left( n, 4 \right) = 2<math> conflicts </math>{\rm Rem} \ \left( n, 2 \right) = 1<math>. |
Therefore, this case has no solution. | Therefore, this case has no solution. | ||
| − | \textbf{Case 4:} | + | </math>\textbf{Case 4:} {\rm Rem} \ \left( n, 6 \right) = 4<math>. |
| − | The condition <math>{\rm Rem} \ \left( n, 6 \right) = 4< | + | The condition </math>{\rm Rem} \ \left( n, 6 \right) = 4<math> implies </math>{\rm Rem} \ \left( n, 2 \right) = 0<math> and </math>{\rm Rem} \ \left( n, 3 \right) = 1<math>. |
| − | Because <math>n< | + | Because </math>n<math> is extra-distinct, </math>{\rm Rem} \ \left( n, l \right)<math> for </math>l \in \left\{ 2, 3, 4 , 5 \right\}<math> is a permutation of </math>\left\{ 0, 1 ,2 , 3 \right\}<math>. |
| − | Because <math>{\rm Rem} \ \left( n, 2 \right) = 0< | + | Because </math>{\rm Rem} \ \left( n, 2 \right) = 0<math>, we must have </math>{\rm Rem} \ \left( n, 4 \right) = 2<math>. Hence, </math>{\rm Rem} \ \left( n, 5 \right) = 3<math>. |
| − | Hence, <math>n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}< | + | Hence, </math>n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}<math>. |
| − | Hence, <math>n \equiv - 2 \pmod{60}< | + | Hence, </math>n \equiv - 2 \pmod{60}<math>. |
| − | We have <math>1000 = 60 \cdot 16 + 40< | + | We have </math>1000 = 60 \cdot 16 + 40<math>. |
| − | Therefore, the number extra-distinct <math>n< | + | Therefore, the number extra-distinct </math>n<math> in this case is 16. |
| − | \textbf{Case 5:} | + | </math>\textbf{Case 5:} {\rm Rem} \ \left( n, 6 \right) = 5<math>. |
| − | The condition <math>{\rm Rem} \ \left( n, 6 \right) = 5< | + | The condition </math>{\rm Rem} \ \left( n, 6 \right) = 5<math> implies </math>{\rm Rem} \ \left( n, 2 \right) = 1<math> and </math>{\rm Rem} \ \left( n, 3 \right) = 2<math>. |
| − | Because <math>n< | + | Because </math>n<math> is extra-distinct, </math>{\rm Rem} \ \left( n, 4 \right)<math> and </math>{\rm Rem} \ \left( n, 5 \right)<math> are two distinct numbers in </math>\left\{ 0, 3, 4 \right\}<math>. |
| − | Because <math>{\rm Rem} \ \left( n, 4 \right) \leq 3< | + | Because </math>{\rm Rem} \ \left( n, 4 \right) \leq 3<math> and </math>n<math> is odd, we have </math>{\rm Rem} \ \left( n, 4 \right) = 3<math>. |
| − | Hence, <math>{\rm Rem} \ \left( n, 5 \right) = 0< | + | Hence, </math>{\rm Rem} \ \left( n, 5 \right) = 0<math> or 4. |
| − | \textbf{Case 5.1:} | + | </math>\textbf{Case 5.1:} {\rm Rem} \ \left( n, 6 \right) = 5<math>, </math>{\rm Rem} \ \left( n, 4 \right) = 3<math>, </math>{\rm Rem} \ \left( n, 5 \right) = 0<math>. |
| − | We have <math>n \equiv 35 \pmod{60}< | + | We have </math>n \equiv 35 \pmod{60}<math>. |
| − | We have <math>1000 = 60 \cdot 16 + 40< | + | We have </math>1000 = 60 \cdot 16 + 40<math>. |
| − | Therefore, the number extra-distinct <math>n< | + | Therefore, the number extra-distinct </math>n<math> in this subcase is 17. |
| − | \textbf{Case 5.2:} | + | </math>\textbf{Case 5.2:} {\rm Rem} \ \left( n, 6 \right) = 5<math>, </math>{\rm Rem} \ \left( n, 4 \right) = 3<math>, </math>{\rm Rem} \ \left( n, 5 \right) = 4<math>. |
| − | <math>n \equiv - 1 \pmod{60}< | + | </math>n \equiv - 1 \pmod{60}<math>. |
| − | We have <math>1000 = 60 \cdot 16 + 40< | + | We have </math>1000 = 60 \cdot 16 + 40<math>. |
| − | Therefore, the number extra-distinct <math>n< | + | 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< | + | Putting all cases together, the total number of extra-distinct </math>n<math> is </math>16 + 17 + 16 = \boxed{\textbf{(049) }}$. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
Revision as of 17:05, 8 February 2023
Unofficial problem statement: Find the number of positive integers from 1 to 1000 that have different mods in mod 2,3,4,5, and 6.
Unofficial Solution: We realize that any such number (mod 2) and (mod 4) must have the same parity, and its values (mod 3) and (mod 6) must have a absolute value difference of 3. Thus the only possibilities for the sequence of mods are 1,2,3,4,5 1,2,3,0,5 and 0,1,2,3,4. Using CRT and summing we get 049.
Solution
.
We have 
. This violates the condition that 
is extra-distinct.
Therefore, this case has no solution.
.
We have 
. This violates the condition that is extra-distinct.
Therefore, this case has no solution.
- \textbf{Case 2:} {\rm Rem} \ \left( n, 6 \right) = 2$.
We have$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 3 \right) = 2
n
\textbf{Case 3:} {\rm Rem} \ \left( n, 6 \right) = 3$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 3
{\rm Rem} \ \left( n, 2 \right) = 1
{\rm Rem} \ \left( n, 3 \right) = 0$.
Because$ (Error compiling LaTeX. Unknown error_msg)n
{\rm Rem} \ \left( n, l \right)
l \in \left\{ 2, 3, 4 \right\}
\left\{ 0, 1 ,2 \right\}
{\rm Rem} \ \left( n, 4 \right) = 2$.
However,$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 4 \right) = 2
{\rm Rem} \ \left( n, 2 \right) = 1
\textbf{Case 4:} {\rm Rem} \ \left( n, 6 \right) = 4$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 4{\rm Rem} \ \left( n, 2 \right) = 0
{\rm Rem} \ \left( n, 3 \right) = 1$.
Because$ (Error compiling LaTeX. Unknown error_msg)n{\rm Rem} \ \left( n, l \right)l \in \left\{ 2, 3, 4 , 5 \right\}\left\{ 0, 1 ,2 , 3 \right\}$.
Because$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 2 \right) = 0
{\rm Rem} \ \left( n, 4 \right) = 2
{\rm Rem} \ \left( n, 5 \right) = 3$.
Hence,$ (Error compiling LaTeX. Unknown error_msg)n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}
n \equiv - 2 \pmod{60}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40
n
\textbf{Case 5:} {\rm Rem} \ \left( n, 6 \right) = 5$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 5{\rm Rem} \ \left( n, 2 \right) = 1{\rm Rem} \ \left( n, 3 \right) = 2$.
Because$ (Error compiling LaTeX. Unknown error_msg)n{\rm Rem} \ \left( n, 4 \right){\rm Rem} \ \left( n, 5 \right)
\left\{ 0, 3, 4 \right\}
{\rm Rem} \ \left( n, 4 \right) \leq 3n
{\rm Rem} \ \left( n, 4 \right) = 3{\rm Rem} \ \left( n, 5 \right) = 0
\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$.
We have$ (Error compiling LaTeX. Unknown error_msg)n \equiv 35 \pmod{60}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40n
\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}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40n$in this subcase is 16.
Putting all cases together, the total number of extra-distinct$ (Error compiling LaTeX. Unknown error_msg)n
16 + 17 + 16 = \boxed{\textbf{(049) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
