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Difference between revisions of "2024 AMC 10A Problems/Problem 3"

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==Video Solution by Dr. David==
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==See also==
 
==See also==
 
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Latest revision as of 19:18, 26 May 2025

Problem

What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes?

$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad\textbf{(E) }13$

Solution 1

Let the requested sum be $S.$ Recall that $2$ is the only even (and the smallest) prime, so $S$ is odd. It follows that the five distinct primes are all odd. The first few odd primes are $3,5,7,11,13,17,19,\ldots.$ We conclude that $S>3+5+7+11+13=39,$ as $39$ is a composite. The next possible value of $S$ is $3+5+7+11+17=43,$ which is a prime. Therefore, we have $S=43,$ and the sum of its digits is $4+3=\boxed{\textbf{(B) }7}.$

~MRENTHUSIASM

Solution 2

We notice that the minimum possible value of the sum of $5$ odd distinct primes is $3 + 5 + 7 + 11 + 13 = 39$, which is not a prime. The smallest prime greater than that is $41$. However, this cannot be written as the sum of $5$ distinct primes, since $15$ is not prime. However, $43$ can be written as $3 + 5 + 7 + 11 + 17 = 43$, so the answer is $4 + 3 = \boxed{\textbf{(B) }7}$

~andliu766

Video Solution

https://youtu.be/l3VrUsZkv8I ~MC

Video Solution by Central Valley Math Circle

https://youtu.be/VJpw7eu7s6M

~mr_mathman

Video Solution by Daily Dose of Math

https://youtu.be/4mf18UuZENw

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=5uruPdMajz7B8jhy&t=307

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/uKXSZyrIOeU?t=900

~IceMatrix

Video Solution by Dr. David

https://youtu.be/Gka4ivb0AEw

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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