Difference between revisions of "1995 AIME Problems/Problem 10"

Revision as of 17:44, 16 August 2020

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 $5$ .

Afterword

Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime, except for 5 itself. Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is $5 mod 42$ . To be exact, $5 * 42 + 5$ .

-jackshi2006

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 $b + 2 \times 42$ is divisible by $5$ and thus $a \leq 2$ . Thus, $n \leq 3 \times 42 = 126 < 215$ .

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

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

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

Our answer is $\boxed{215}$ .

@@ Line 28: / Line 28: @@
 ====Afterword====
-Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime (except for 5 itself). Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is <math>5 mod 42</math>. To be exact, <math>5 * 42 + 5</math>.
+Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime, except for 5 itself. Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is <math>5 mod 42</math>. To be exact, <math>5 * 42 + 5</math>.
 -jackshi2006

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

Page

Toolbox

Search