2024 AMC 10B Problems/Problem 7

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 $19$.

~ILoveMath31415926535

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions

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