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

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What is the sum of the digits of the smallest prime that can be written as a sum of <math>5</math> distinct primes?
 
What is the sum of the digits of the smallest prime that can be written as a sum of <math>5</math> distinct primes?
  
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }13</math>
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<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad\textbf{(E) }13</math>
  
== Solution ==
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== Solution 1==
Recall that <math>2</math> is the only even prime.
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Let the requested sum be <math>S.</math> Recall that <math>2</math> is the only even (and the smallest) prime, so <math>S</math> is odd. It follows that the five distinct primes are all odd. The first few odd primes are <math>3,5,7,11,13,17,19,\ldots.</math> We conclude that <math>S>3+5+7+11+13=39,</math> as <math>39</math> is a composite. The next possible value of <math>S</math> is <math>3+5+7+11+17=43,</math> which is a prime. Therefore, we have <math>S=43,</math> and the sum of its digits is <math>4+3=\boxed{\textbf{(B) }7}.</math>
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~MRENTHUSIASM
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== Solution 2 ==
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We notice that the minimum possible value of the sum of <math>5</math> odd distinct primes is <math>3 + 5 + 7 + 11 + 13 = 39</math>, which is not a prime. The smallest prime greater than that is <math>41</math>. However, this cannot be written as the sum of <math>5</math> distinct primes, since <math>15</math> is not prime. However, <math>43</math> can be written as <math>3 + 5 + 7 + 11 + 17 = 43</math>, so the answer is <math>4 + 3 = \boxed{\textbf{(B) }7}</math>
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~andliu766
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/4mf18UuZENw
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~Thesmartgreekmathdude
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== Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=5uruPdMajz7B8jhy&t=307
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=6SQ74nt3ynw
  
Let the requested sum be <math>S.</math> Clearly, <math>S</math> is odd. It follows that the five distinct primes are all odd.
 
  
 
==See also==
 
==See also==
{{AMC10 box|year=2024|ab=A|before=2|num-a=4}}
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{{AMC10 box|year=2024|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:03, 20 November 2024

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 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


See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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

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