Difference between revisions of "2024 AMC 10A Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | 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. | + | 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> from which <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> |
| − | + | ~MRENTHUSIASM | |
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|before=2|num-a=4}} | {{AMC10 box|year=2024|ab=A|before=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:34, 8 November 2024
Problem
What is the sum of the digits of the smallest prime that can be written as a sum of 
distinct primes?
Solution
Let the requested sum be 
Recall that 
is the only even (and the smallest) prime, so 
is odd. It follows that the five distinct primes are all odd. The first few odd primes are 
from which 
as 
is a composite. The next possible value of is 
which is a prime. Therefore, we have 
and the sum of its digits is 
~MRENTHUSIASM
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
