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

Latest revision as of 15:03, 12 August 2025

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (Given more advanced knowledge)
6 Solution 5 (Basic Mod Almost same as sol one)
7 🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)
8 Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
9 Video Solution 2 by SpreadTheMathLove
10 See also

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

Solution 5 (Basic Mod Almost same as sol one)

$7^{2024} * (1+7+49) == 0 (mod 19)$ $57 == 0 mod 19$

so it is just $\boxed{\textbf{(A) } 0}$ .

~Cheerfulfrog

🎥✨ 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 36: / Line 36: @@
 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>.
-~ILoveMath31415926535
+==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
+==Solution 5 (Basic Mod Almost same as sol one)==
+<math>7^{2024} * (1+7+49) == 0 (mod 19)</math>
+<math>57 == 0 mod 19</math>
+so it is just <math>\boxed{\textbf{(A) } 0}</math>.
+~Cheerfulfrog
 ==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)==

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