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

(→‎Solution 1)
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Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>.
 
Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>.
 +
 +
==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.
 +
 +
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>
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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]]
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:39, 14 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 $57$ is divisible by $19$, 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

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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