View source for 2023 AIME I Problems/Problem 7

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

==Solution 1==

<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>

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>

<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.

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

~mathboy100

==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>.

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.

<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.
Therefore, this case has no solution.

<math>\textbf{Case 2:  } {\rm Rem} \ \left( n, 6 \right) = 2</math>.

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.

<math>\textbf{Case 3:  } {\rm Rem} \ \left( n, 6 \right) = 3</math>.

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</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</math>.

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.

<math>\textbf{Case 4:  } {\rm Rem} \ \left( n, 6 \right) = 4</math>.

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</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</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)}</math>.
Hence, <math>n \equiv - 2 \pmod{60}</math>.

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

<math>\textbf{Case 5:  } {\rm Rem} \ \left( n, 6 \right) = 5</math>.

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</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</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</math> or 4.

<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}</math>.

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

<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>.

We have <math>1000 = 60 \cdot 16 + 40</math>.
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{049}</math>.

~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>

'''vladimir.shelomovskii@gmail.com, vvsss'''

==Video Solution==
https://youtu.be/8oOik9d1fWM

~MathProblemSolvingSkills.com

==See also==
{{AIME box|year=2023|num-b=6|num-a=8|n=I}}

[[Category:Intermediate Number Theory Problems]]
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