2022 AMC 10A Problems/Problem 19
Contents
Problem
Define as the least common multiple of all the integers from to inclusive. There is a unique integer such that What is the remainder when is divided by ?
Solution 1
Notice that contains the highest power of every prime below since higher primes cannot divide . Thus, .
When writing the sum under a common fraction, we multiply the denominators by divided by each denominator. However, since is a multiple of , all terms will be a multiple of until we divide out , and the only term that will do this is . Thus, the remainder of all other terms when divided by will be , so the problem is essentially asking us what the remainder of divided by is. This is equivalent to finding the remainder of divided by .
We use modular arithmetic to simplify our answer:
This is congruent to .
Evaluating, we get: Therefore the remainder is .
~KingRavi
~mathboy282
~Scarletsyc
~wangzrpi
Solution 2
As in solution 1, we express the LHS as a sum under one common denominator. We note that
Now, we have . We'd like to find so we can evaluate our expression Since don't have a factor of in their denominators, and since is a multiple of multiplying each of those terms and adding them will get a multiple of , that result is Thus, we only need to consider Proceed with solution to get .
~sirswagger21
Solution 3
Using Wolstenholmes' Theorem, we can rewrite as (for some ). Adding the to , we get .
Now we have and we want . We find that . Taking and multiplying, we get .
Applying Wilson's Theorem on and reducing, we simplify the congruence to . Now we proceed with Solution 1 and find that , so our answer is .
~kn07
Video Solution (⚡️Under 3 min⚡️)
~Education, the Study of Everything
Video Solution By ThePuzzlr
~ MathIsChess
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.