Difference between revisions of "1995 AIME Problems/Problem 10"
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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. | 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. |
Revision as of 17:26, 30 March 2023
Contents
[hide]Problem
What is the largest positive integer that is not the sum of a positive integral multiple of and a positive composite integer?
Solution
The requested number must be a prime number (or 1). Also, every number that is a multiple of 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 and the numbers that are multiples of greater than them, until they reach a composite number.
is the greatest number in the list, so it is the answer. Note that considering would have shortened the search, since , and so within numbers at least one must be divisible by .
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 . Specifically, .
-jackshi2006
Solution 2
Let our answer be . Write , where are positive integers and . Then note that are all primes.
If is , then because is the only prime divisible by . We get as our largest possibility in this case.
If is , then is divisible by and thus . Thus, .
If is , then is divisible by and thus . Thus, .
If is , then is divisible by and thus . Thus, .
If is , then is divisible by and thus . Thus, .
Our answer is .
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.