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Difference between revisions of "2012 AIME II Problems/Problem 12"
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== Solution ==
 
== Solution ==
  
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We see that a number <math>n</math> is <math>p</math>-safe if and only if the residue of <math>n \mod p</math> is greater than <math>2</math> and less than <math>p-2</math>; thus, there are <math>p-5</math> residues <math>\mod p</math> that a <math>p</math>-safe number can have. Therefore, a number <math>n</math> satisfying the conditions of the problem can have <math>2</math> different residues <math>\mod 7</math>, <math>6</math> different residues <math>\mod 11</math>, and <math>8</math> different residues <math>\mod 13</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>10007</math> and <math>10006</math>, so there remain <math>\fbox{958}</math> values satisfying the conditions of the problem.
+
We see that a number <math>n</math> is <math>p</math>-safe if and only if the residue of <math>n \mod p</math> is greater than <math>2</math> and less than <math>p-2</math>; thus, there are <math>p-5</math> residues <math>\mod p</math> that a <math>p</math>-safe number can have. Therefore, a number <math>n</math> satisfying the conditions of the problem can have <math>2</math> different residues <math>\mod 7</math>, <math>6</math> different residues <math>\mod 11</math>, and <math>8</math> different residues <math>\mod 13</math>. The Chinese Remainder Theorem states that for a number <math>x</math> that is
 +
<math>a (mod b)</math>
 +
<math>c (mod d)</math>
 +
<math>e (mod f)</math>
 +
has one solution if <math>gcd(b,d,f)=1. For example, in our case, the number </math>n<math> can be:
 +
</math>3 (mod 7)<math>
 +
</math>3 (mod 11)<math>
 +
</math>7 (mod 13)<math>
 +
so since gcd(7,11,13)=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>10007<math> and </math>10006<math>, so there remain </math>\fbox{958}$ values satisfying the conditions of the problem.
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:11, 15 February 2015

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)$$ (Error compiling LaTeX. Unknown error_msg)3 (mod 11)$$ (Error compiling LaTeX. Unknown error_msg)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,$ (Error compiling LaTeX. Unknown error_msg)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$0 \le n < 10010$. 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$10007$and$10006$, so there remain$\fbox{958}$ values satisfying the conditions of the problem.

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

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