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

Revision as of 21:28, 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 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

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

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 3

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 961-1000 will act the same as 1-40, where we have 1 extra-distinct number. 48 + 1 = \boxed{\textbf{(49) }}$

@@ Line 84: / Line 84: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution 3==
+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. <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. 48 + 1 =  \boxed{\textbf{(49) }}$
 ==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
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