Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AIME II Problems/Problem 12"
(→‎Solution)
 
(4 intermediate revisions by 4 users not shown)
Line 8: Line 8:
 
<math>c</math> (mod d)
 
<math>c</math> (mod d)
 
<math>e</math> (mod f)
 
<math>e</math> (mod f)
−
has one solution if <math>gcd(b,d,f)=1</math>. For example, in our case, the number <math>n</math> can be:
+
has one solution if <math>\gcd(b,d,f)=1</math>. For example, in our case, the number <math>n</math> can be:
 
3 (mod 7)
 
3 (mod 7)
 
3 (mod 11)
 
3 (mod 11)
 
7 (mod 13)
 
7 (mod 13)
−
so since <math>gcd(7,11,13)</math>=1, there is 1 solution for n for this case of residues of <math>n</math>.
+
so since <math>\gcd(7,11,13)</math>=1, there is 1 solution for n for this case of residues of <math>n</math>.
  
−
This means that by the Chinese Remainder Theorem, <math>n</math> can have <math>2\cdot 6 \cdot 8 = 96</math> different residues mod <math>7 \cdot 11 \cdot 13 = 1001</math>. Thus, there are <math>960</math> values of <math>n</math> satisfying the conditions in the range <math>0 \le n < 10010</math>. However, we must now remove any values greater than <math>10000</math> that satisfy the conditions. By checking residues, we easily see that the only such values are <math>10006</math> and <math>10007</math>, so there remain <math>\fbox{958}</math> values satisfying the conditions of the problem.
+
This means that by the Chinese Remainder Theorem, <math>n</math> can have <math>2\cdot 6 \cdot 8 = 96</math> different residues mod <math>7 \cdot 11 \cdot 13 = 1001</math>. Thus, there are <math>960</math> values of <math>n</math> satisfying the conditions in the range <math>0 < n \le 10010</math>. However, we must now remove any values greater than <math>10000</math> that satisfy the conditions. By checking residues, we easily see that the only such values are <math>10006</math> and <math>10007</math>, so there remain <math>\fbox{958}</math> values satisfying the conditions of the problem.
 +
 
 +
 
 +
*We can also say <math>\frac{2}{7}</math> of all numbers are 7-safe, <math>\frac{6}{11}</math> of all numbers are 11-safe, and <math>\frac{8}{13}</math> of all numbers are 13-safe. We can multiply these to get that <math>\frac{96}{1001}</math> of all numbers are simultaneously 7-safe, 11-safe, and 13-safe.
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2012|n=II|num-b=11|num-a=13}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:52, 12 June 2024

Problem 12

For a positive integer $p$, define the positive integer $n$ to be $p$-safe if $n$ differs in absolute value by more than $2$ from all multiples of $p$. For example, the set of $10$-safe numbers is $\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}$. Find the number of positive integers less than or equal to $10,000$ which are simultaneously $7$-safe, $11$-safe, and $13$-safe.

Solution

We see that a number $n$ is $p$-safe if and only if the residue of $n \mod p$ is greater than $2$ and less than $p-2$; thus, there are $p-5$ residues $\mod p$ that a $p$-safe number can have. Therefore, a number $n$ satisfying the conditions of the problem can have $2$ different residues $\mod 7$, $6$ different residues $\mod 11$, and $8$ different residues $\mod 13$. The Chinese Remainder Theorem states that for a number $x$ that is $a$ (mod b) $c$ (mod d) $e$ (mod f) has one solution if $\gcd(b,d,f)=1$. For example, in our case, the number $n$ can be: 3 (mod 7) 3 (mod 11) 7 (mod 13) so since $\gcd(7,11,13)$=1, there is 1 solution for n for this case of residues of $n$.

This means that by the Chinese Remainder Theorem, $n$ can have $2\cdot 6 \cdot 8 = 96$ different residues mod $7 \cdot 11 \cdot 13 = 1001$. Thus, there are $960$ values of $n$ satisfying the conditions in the range $0 < n \le 10010$. However, we must now remove any values greater than $10000$ that satisfy the conditions. By checking residues, we easily see that the only such values are $10006$ and $10007$, so there remain $\fbox{958}$ values satisfying the conditions of the problem.


  • We can also say $\frac{2}{7}$ of all numbers are 7-safe, $\frac{6}{11}$ of all numbers are 11-safe, and $\frac{8}{13}$ of all numbers are 13-safe. We can multiply these to get that $\frac{96}{1001}$ of all numbers are simultaneously 7-safe, 11-safe, and 13-safe.

See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_12&oldid=220403"