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 "1995 AIME Problems/Problem 10"
m (→‎Second Solution)
Line 28: Line 28:
 
== Second Solution ==
 
== Second Solution ==
  
−
Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> 0 <= b < 42 </math>. Then note that <math> b, b + 42, ... , b + 42(a-1) </math> are all primes.  
+
Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> 0 \leq b < 42 </math>. Then note that <math> b, b + 42, ... , b + 42(a-1) </math> are all primes.  
  
 
If <math>b</math> is <math>0\mod{5}</math>, then <math>b = 5</math> because 5 is the only prime divisible by 5. We get <math> n = 215</math> as our largest possibility in this case.  
 
If <math>b</math> is <math>0\mod{5}</math>, then <math>b = 5</math> because 5 is the only prime divisible by 5. We get <math> n = 215</math> as our largest possibility in this case.  

Revision as of 15:30, 27 February 2018

Problem

What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer?

Solution

The requested number $\mod {42}$ must be a prime number. Also, every number that is a multiple of $42$ greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to $42$ and the numbers that are multiples of $42$ greater than them, until they reach a composite number.

\[\begin{tabular}{|r||r|r|r|r|r|} \hline 2&44&&&& \\ 3&45&&&& \\ 5&47&89&131&173&215 \\ 7&49&&&& \\ 11&53&95&&& \\ 13&55&&&& \\ 17&59&101&143&& \\ 19&61&103&145&& \\ 23&65&&&& \\ 29&71&113&155&& \\ 31&73&115&&& \\ 37&79&121&&& \\ 41&83&125&&& \\ \hline \end{tabular}\]

$\boxed{215}$ is the greatest number in the list, so it is the answer. Note that considering $\mod {5}$ would have shortened the search, since $\text{gcd}(5,42)=1$, and so within $5$ numbers at least one must be divisible by $5$.

Second Solution

Let our answer be $n$. Write $n = 42a + b$, where $a, b$ are positive integers and $0 \leq b < 42$. Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.

If $b$ is $0\mod{5}$, then $b = 5$ because 5 is the only prime divisible by 5. We get $n = 215$ as our largest possibility in this case.

If $b$ is $1\mod{5}$, then $b + 2 * 42$ is divisible by 5 and thus $a <= 2$. Thus, $n <= 3*42 = 126 < 215$.

If $b$ is $2\mod{5}$, then $b + 4 * 42$ is divisible by 5 and thus $a <= 4$. Thus, $n <= 5*42 = 210 < 215$.

If $b$ is $3\mod{5}$, then $b + 1 * 42$ is divisible by 5 and thus $a = 1$. Thus, $n <= 2*42 = 84 < 215$.

If $b$ is $4\mod{5}$, then $b + 3 * 42$ is divisible by 5 and thus $a <= 3$. Thus, $n <= 4*42 = 168 < 215$.

Our answer is $\boxed{215}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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=1995_AIME_Problems/Problem_10&oldid=92428"