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Difference between revisions of "2012 AMC 8 Problems/Problem 18"
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==Solution==
 
==Solution==
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The problem states that the answer cannot be a perfect square or have prime factors less than <math> 50 </math>. Therefore, the answer will be the product of at least two different primes grater than <math> 50 </math>. The two smallest primes greater than <math> 50 </math> are <math> 53 </math> and <math> 59 </math>. Multiplying these two primes, we obtain the number <math> 3127 </math>, which is also the smallest number on the list of answer choices. So we are done, and the answer is <math> \boxed{\textbf{(A)}\ 3127}</math>.
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The problem states that the answer cannot be a perfect square or have prime factors less than <math>50</math>. Therefore, the answer will be the product of at least two different primes greater than <math>50</math>. The two smallest primes greater than <math>50</math> are <math>53</math> and <math>59</math>. Multiplying these two primes, we obtain the number <math>3127</math>, which is also the smallest number on the list of answer choices.  
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 +
So we are done, and the answer is <math>\boxed{\textbf{(A)}\ 3127}</math>.
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 +
== Video Solutions ==
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https://youtu.be/HISL2-N5NVg?t=526
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~ pi_is_3.14159
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https://youtu.be/qBXOgsZlCg4 ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=17|num-a=19}}
 
{{AMC8 box|year=2012|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 09:23, 21 December 2024

Problem

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?

$\textbf{(A)}\hspace{.05in}3127\qquad\textbf{(B)}\hspace{.05in}3133\qquad\textbf{(C)}\hspace{.05in}3137\qquad\textbf{(D)}\hspace{.05in}3139\qquad\textbf{(E)}\hspace{.05in}3149$

Solution

The problem states that the answer cannot be a perfect square or have prime factors less than $50$. Therefore, the answer will be the product of at least two different primes greater than $50$. The two smallest primes greater than $50$ are $53$ and $59$. Multiplying these two primes, we obtain the number $3127$, which is also the smallest number on the list of answer choices.

So we are done, and the answer is $\boxed{\textbf{(A)}\ 3127}$.

Video Solutions

https://youtu.be/HISL2-N5NVg?t=526

~ pi_is_3.14159

https://youtu.be/qBXOgsZlCg4 ~savannahsolver

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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