Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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Using Wolstenholmes' Theorem, we can rewrite <math>1 + \frac{1}{2} \dots + \frac{1}{16}</math> as <math>\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}</math> (for some <math>n \in \mathbb{Z}</math>). Adding the <math>\frac{1}{17}</math> to <math>\frac{17^2 n}{16!}</math>, we get <math>\frac{17^3 n + 16!}{17!}</math>. | Using Wolstenholmes' Theorem, we can rewrite <math>1 + \frac{1}{2} \dots + \frac{1}{16}</math> as <math>\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}</math> (for some <math>n \in \mathbb{Z}</math>). Adding the <math>\frac{1}{17}</math> to <math>\frac{17^2 n}{16!}</math>, we get <math>\frac{17^3 n + 16!}{17!}</math>. | ||
Now we have <math>\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}</math> and we want <math>h \pmod{17}</math>. We find that <math>\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h</math>. Taking <math>\pmod{17}</math> and multiplying, we get <math>L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}</math>. | Now we have <math>\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}</math> and we want <math>h \pmod{17}</math>. We find that <math>\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h</math>. Taking <math>\pmod{17}</math> and multiplying, we get <math>L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}</math>. | ||
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Applying Wilson's Theorem on <math>16!</math> and reducing, we simplify the congruence to <math>L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}</math>. Now we proceed with Solution 1 and find that <math>L_{16} \equiv 5 \pmod{17}</math>, so our answer is <math>\boxed{\textbf{(C) }5}</math>. | Applying Wilson's Theorem on <math>16!</math> and reducing, we simplify the congruence to <math>L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}</math>. Now we proceed with Solution 1 and find that <math>L_{16} \equiv 5 \pmod{17}</math>, so our answer is <math>\boxed{\textbf{(C) }5}</math>. | ||
Revision as of 17:46, 10 December 2022
Contents
[hide]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 . 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
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 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.