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Difference between revisions of "2024 AMC 10B Problems/Problem 7"

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==Problem==
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What is the remainder when <math>7^{2024}+7^{2025}+7^{2026}</math> is divided by <math>19</math>?
  
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<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 18</math>
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==Solution 1==
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We can factor the expression as
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<cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath>
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Note that <math>57=19\cdot3</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>.
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==Solution 2==
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If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem.
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Since <math>7^3\equiv1\pmod{19}</math>, the powers of <math>7</math> repeat every three terms:
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<cmath>7^1\equiv7\pmod{19}</cmath>
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<cmath>7^2\equiv11\pmod{19}</cmath>
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<cmath>7^3\equiv1\pmod{19}</cmath>
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The fact that <math>2024\equiv2\pmod3</math>, <math>2025\equiv0\pmod3</math>, and <math>2026\equiv1\pmod3</math> implies that <math>7^{2024}+7^{2025}+7^{2026}\equiv11+1+7\equiv19 \equiv0\pmod{19}</math>.
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~[[User:Bloggish|Bloggish]]
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==Solution 3==
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We start the same as solution 2, and find that:
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<cmath>7^1\equiv7\pmod{19}</cmath>
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<cmath>7^2\equiv11\pmod{19}</cmath>
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<cmath>7^3\equiv1\pmod{19}</cmath>
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We know that for <math>2024</math>, <math>2025</math>, and <math>2026</math>, because there are three terms, we can just add them up. <math>1 + 7 + 11 = 19</math>, which is <math>0</math> mod <math>19</math>.
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==Solution 4 (Given more advanced knowledge)==
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By Fermat's Little Theorem (FLT), we know that <cmath>7^{18}\equiv1\pmod{19}</cmath> Then its order must divide <math>18</math>. Trying simple values we try and succeed: <cmath>7^3\equiv1\pmod{19}</cmath>
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So the expression is equivalent to <math>1+7+49\pmod{19}</math>, which gives <math>\boxed{\textbf{(A) } 0}</math> when divided by 19.
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~xHypotenuse
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==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)==
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https://youtu.be/T_QESWAKUUk?si=5euBbKNMaYBROuTV&t=100
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/QLziG_2e7CY?feature=shared
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~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 00:52, 21 November 2024

Problem

What is the remainder when $7^{2024}+7^{2025}+7^{2026}$ is divided by $19$?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 18$

Solution 1

We can factor the expression as

\[7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).\]

Note that $57=19\cdot3$, this expression is actually divisible by 19. The answer is $\boxed{\textbf{(A) } 0}$.

Solution 2

If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem.

Since $7^3\equiv1\pmod{19}$, the powers of $7$ repeat every three terms:

\[7^1\equiv7\pmod{19}\] \[7^2\equiv11\pmod{19}\] \[7^3\equiv1\pmod{19}\]

The fact that $2024\equiv2\pmod3$, $2025\equiv0\pmod3$, and $2026\equiv1\pmod3$ implies that $7^{2024}+7^{2025}+7^{2026}\equiv11+1+7\equiv19 \equiv0\pmod{19}$.

~Bloggish

Solution 3

We start the same as solution 2, and find that:

\[7^1\equiv7\pmod{19}\] \[7^2\equiv11\pmod{19}\] \[7^3\equiv1\pmod{19}\]

We know that for $2024$, $2025$, and $2026$, because there are three terms, we can just add them up. $1 + 7 + 11 = 19$, which is $0$ mod $19$.


Solution 4 (Given more advanced knowledge)

By Fermat's Little Theorem (FLT), we know that \[7^{18}\equiv1\pmod{19}\] Then its order must divide $18$. Trying simple values we try and succeed: \[7^3\equiv1\pmod{19}\]


So the expression is equivalent to $1+7+49\pmod{19}$, which gives $\boxed{\textbf{(A) } 0}$ when divided by 19.


~xHypotenuse

🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)

https://youtu.be/T_QESWAKUUk?si=5euBbKNMaYBROuTV&t=100

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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