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

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

@@ Line 1: / Line 1: @@
-Alice wants to play Uno with Bob on the Xbox but Bob says he has the oldest xbox known to man. What is the probability that Bob does not have Uno
+==Problem==
+What is the remainder when <math>7^{2024}+7^{2025}+7^{2026}</math> is divided by <math>19</math>?
+<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 18</math>
+==Solution 1==
+We can factor the expression as
+<cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath>
+Note that <math>57=19\cdot3</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>.
+==Solution 2==
+If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem.
+Since <math>7^3\equiv1\pmod{19}</math>, the powers of <math>7</math> repeat every three terms:
+<cmath>7^1\equiv7\pmod{19}</cmath>
+<cmath>7^2\equiv11\pmod{19}</cmath>
+<cmath>7^3\equiv1\pmod{19}</cmath>
+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>.
+~[[User:Bloggish|Bloggish]]
+==Solution 3==
+We start the same as solution 2, and find that:
+<cmath>7^1\equiv7\pmod{19}</cmath>
+<cmath>7^2\equiv11\pmod{19}</cmath>
+<cmath>7^3\equiv1\pmod{19}</cmath>
+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>.
+==Solution 4 (Given more advanced knowledge)==
+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>
+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.
+~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==
+{{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}}
+{{MAA Notice}}

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