Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
− | 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>. | + | Notice that <math>L_{17}</math> contains the highest power of every prime below <math>17</math> since higher primes cannot divide <math>L_{17}</math>. Thus, <math>L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17</math>. |
− | 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} = | + | 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} = L_{16}</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>. |
We use modular arithmetic to simplify our answer: | We use modular arithmetic to simplify our answer: | ||
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~Scarletsyc | ~Scarletsyc | ||
+ | |||
+ | ~wangzrpi | ||
== Solution 2 == | == Solution 2 == | ||
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~kn07 | ~kn07 | ||
+ | |||
+ | |||
+ | ==Video Solution (⚡️3 min⚡️)== | ||
+ | https://youtu.be/3g39lB6XLAE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== Video Solution By ThePuzzlr == | == Video Solution By ThePuzzlr == |
Latest revision as of 20:38, 29 October 2023
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 (⚡️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.