Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \cong 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</cmath> | Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \cong 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</cmath> | ||
| − | <cmath> \cong 42 \cdot 11 cdot 13 \pmod{17} \cong 8 \cdot 11 \cdot 13 \pmod{17}</cmath> | + | <cmath> \cong 42 \cdot 11 \cdot 13 \pmod{17} \cong 8 \cdot 11 \cdot 13 \pmod{17}</cmath> |
<cmath> \cong 88 \cdot 13 \pmod{17} \cong 3 \cdot 13 \pmod{17}</cmath> | <cmath> \cong 88 \cdot 13 \pmod{17} \cong 3 \cdot 13 \pmod{17}</cmath> | ||
<cmath>\cong 39 \pmod{17} \cong 5\pmod{17}</cmath> | <cmath>\cong 39 \pmod{17} \cong 5\pmod{17}</cmath> | ||
| − | Therefore the remainder is <math>5</math> and the answer is <math>\boxed{ | + | Therefore the remainder is <math>5</math> and the answer is <math>\boxed{C}</math> |
Revision as of 13:30, 12 November 2022
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
Notice that 
contains the highest power of every prime below . 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: ![\[-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \cong 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}\]](http://latex.artofproblemsolving.com/e/0/4/e0456a63e7ab118e7836d5a737ef89e63ecfe5fc.png)
![\[\cong 42 \cdot 11 \cdot 13 \pmod{17} \cong 8 \cdot 11 \cdot 13 \pmod{17}\]](http://latex.artofproblemsolving.com/b/d/2/bd26c85a68ffcb00522fc994d6429e655381f608.png)
![\[\cong 88 \cdot 13 \pmod{17} \cong 3 \cdot 13 \pmod{17}\]](http://latex.artofproblemsolving.com/c/3/a/c3a4d2bbc368de5ab4296b295a069cea117c7335.png)
![\[\cong 39 \pmod{17} \cong 5\pmod{17}\]](http://latex.artofproblemsolving.com/0/8/3/083079a27a2006775963764c7d928ec4593009de.png)
Therefore the remainder is 
and the answer is 
~KingRavi
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. 
