Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math> | ||
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| + | ==Solution== | ||
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| + | Notice that <math>L_{17}</math> contains the highest power of every prime below <math>17</math>. Thus, <math>L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17</math>. | ||
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| + | When writing the sum under a common fraction, we multiply the denominators by <math>L_{17}</math> divided by each denominator. However, since <math>L_{17}</math> is a multiple of <math>17</math>, all terms will be a multiple of <math>17</math> until we divide out <math>17</math>, and the only term that will do this is <math>\frac{1}{17}</math>. Thus, the remainder of all other terms when divided by <math>17</math> will be <math>0</math>, so the problem is essentially asking us what the remainder of <math>\frac{L_{17}}{17}</math> divided by <math>17</math> is. This is equivalent to finding the remainder of <math>16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13</math> divided by <math>17</math>. | ||
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| + | We use modular arithmetic to simplify our answer: | ||
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| + | This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 (mod 17)</math> | ||
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| + | Solution in Progress | ||
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| + | ~KingRavi | ||
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== Video Solution By ThePuzzlr == | == Video Solution By ThePuzzlr == | ||
Revision as of 13:26, 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 
Solution in Progress
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
