sum of 7th powers

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Source: 2024 AMC 10B #7

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872 posts

#1 h Nov 13, 2024, 5:20 PM

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

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

This post has been edited 1 time. Last edited by LauraZed, Nov 13, 2024, 5:28 PM
Reason: changing to official wording, adding answer options, and filling out source field

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632 posts

#2 h Nov 13, 2024, 5:21 PM

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answer is $0$ ; factor out a $7^{2024}$ , then $7^2+7+1\equiv 0 \pmod {19}$

i did what @below did in contest

This post has been edited 4 times. Last edited by gladIasked, Nov 13, 2024, 5:24 PM

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872 posts

#3 h Nov 13, 2024, 5:21 PM

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what i did was $7^3\equiv 1\pmod{19}$ lol? so it's $7^2+7+1\equiv 0\pmod{19}$

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2989 posts

#4 h Nov 13, 2024, 5:39 PM • 1 Y

Y by beastEJ

cycled 7,11,1 mod 19, and because 2024 == 2 mod 3, we have mod 19 to be 11+1+7=19 mod 19 ==0 (A)

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635 posts

#5 h Nov 13, 2024, 5:54 PM

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its just 0, find 7^4 mod 19, and work from there

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pingpongmerrily

3518 posts

pingpongmerrily

#6 h Nov 13, 2024, 5:57 PM • 1 Y

Y by Sedro

MathRook7817 wrote:

its just 0, find 7^4 mod 19, and work from there

not necessary to bash that, just factor out $7^2+7+1=57$ which is divisible by $19$ ans is 0=A

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115 posts

#7 h Nov 13, 2024, 6:05 PM • 1 Y

Y by wangzrpi

mathboy282 wrote:

cycled 7,11,1 mod 19, and because 2024 == 2 mod 3, we have mod 19 to be 11+1+7=19 mod 19 ==0 (A)

Note that 2024, 2025, and 2026 are 3 consecutive integers and the mod 19 residues of powers of 7 cycle back every 3 numbers. So you can just add the possibilities (residue of 1+11+7 mod 19) so answer is 0

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mathprodigy2011

244 posts

mathprodigy2011

#8 h Nov 13, 2024, 6:23 PM

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omg I thought that said 10b #17. and if it was 17 I would've flipped cuz I took 12b

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739 posts

#9 h Nov 13, 2024, 6:25 PM

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I misread the 19 as a 17 and got clinical depression during the test

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115 posts

#10 h Nov 13, 2024, 7:05 PM

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This should have been named "Lucky Sevens 2024"

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548 posts

#11 h Nov 13, 2024, 8:27 PM

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Kinda ez for a P.7. $7^{2024}+7^{2025}+7^{2026}=7^{2024}(1+7+7^2)=7^{2024}(1+7+7^2)=7^{2024}(57)=7^{2024}(3)(19).$ So $7^{2024}+7^{2025}+7^{2026}$ is divisible by $19$ . Therefore the answer is $\boxed{A}$ .

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525 posts

#12 h Nov 13, 2024, 9:16 PM

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doing some quick modulo math we find 7^1=7 7^2=11 and 7^3=1 mod 19
notice that since 2024 2025 and 2026 are consecutive and 7^x mod 19 loops every 3 numbers we can simply add 7+1+11=19=0 mod 19

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225 posts

#13 h Nov 13, 2024, 9:21 PM

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The powers of seven look between 7,1,11 in some order, so the sum is 19, which is 0(mod 19), or A

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161 posts

#14 h Nov 13, 2024, 9:24 PM

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gracemoon124 wrote:

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 \qquad this was in amc 10 b $

also i just did it

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1725 posts

#15 h Nov 13, 2024, 9:30 PM

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I'm like the only person in my school who didn't silly this somehow

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pingpongmerrily

3518 posts

pingpongmerrily

#16 h Nov 13, 2024, 9:34 PM

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how do you silly this do kids not learn to factor nowadays? what are we coming to? the world is ending!!!

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548 posts

#17 h Nov 13, 2024, 9:41 PM

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fr :skull: silling easy probs with no traps and annoying bashing is just a skill issue. Also, why does everyone use mod. They arent gonna test us on mod on a prob 7.

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1725 posts

#18 h Nov 13, 2024, 9:43 PM

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pingpongmerrily wrote:

how do you silly this do kids not learn to factor nowadays? what are we coming to? the world is ending!!!

Idk man
There's this one guy who got like 132 or smth around that range and this was one of his sillies

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192 posts

#19 h Nov 13, 2024, 9:44 PM

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I used both finding a pattern and factoring to double check my answer.

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1466 posts

#20 h Nov 15, 2024, 11:42 PM

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factor it out
$7^{2024}(1+7+49)$
$1+7+49=57$
$57$ IS divisible by $19$
so our answer should be $\boxed{A:0}$

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2833 posts

#21 h Nov 16, 2024, 12:20 AM • 2 Y

Y by LostDreams, anticodon

Solution

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