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==Problem== | ==Problem== | ||
− | Define <math>L_n</math> as the least common multiple of all the integers from <math>1</math> to <math>n</math> inclusive. There is a unique integer <math>h</math> such that | + | Define <math>L_n</math> as the least common multiple of all the integers from <math>1</math> to <math>n</math> inclusive. There is a unique integer <math>h</math> such that |
− | + | <cmath>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}</cmath> | |
− | < | ||
− | |||
What is the remainder when <math>h</math> is divided by <math>17</math>? | What is the remainder when <math>h</math> is divided by <math>17</math>? | ||
<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> | ||
− | ==Solution== | + | ==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}</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>. | + | 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: | ||
− | This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math> | + | This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math>. |
Evaluating, we get: | Evaluating, we get: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | - | + | (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ |
− | &\equiv | + | &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ |
− | &\equiv | + | &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ |
− | &\equiv | + | &\equiv 2 \cdot 11 \pmod{17} \\ |
− | |||
− | |||
&\equiv 5\pmod{17} | &\equiv 5\pmod{17} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Therefore the remainder is <math>\boxed{\textbf{(C) } 5}</math>. | ||
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ~Scarletsyc | ||
+ | |||
+ | ~wangzrpi | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | As in solution 1, we express the LHS as a sum under one common denominator. We note that <cmath>\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}</cmath> | ||
+ | |||
+ | Now, we have <math>h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)</math>. We'd like to find <math>h \pmod{17},</math> so we can evaluate our expression <math>\pmod{17}.</math> Since <math>\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}</math> don't have a factor of <math>17</math> in their denominators, and since <math>L_{17}</math> is a multiple of <math>17,</math> multiplying each of those terms and adding them will get a multiple of <math>17.</math> <math>\pmod{17}</math>, that result is <math>0.</math> Thus, we only need to consider <math>L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.</math> Proceed with solution <math>1</math> to get <math>\boxed{\textbf{(C) }5}</math>. | ||
+ | |||
+ | ~sirswagger21 | ||
+ | |||
+ | == Solution 3 == | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~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.