Difference between revisions of "2012 AMC 8 Problems/Problem 18"

(See Also)
m
 
(4 intermediate revisions by 3 users not shown)
Line 9: Line 9:
 
So we are done, and the answer is <math>\boxed{\textbf{(A)}\ 3127}</math>.
 
So we are done, and the answer is <math>\boxed{\textbf{(A)}\ 3127}</math>.
  
 
+
== Video Solutions ==
== Video Solution ==
 
 
https://youtu.be/HISL2-N5NVg?t=526
 
https://youtu.be/HISL2-N5NVg?t=526
  

Latest revision as of 12:58, 10 November 2023

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_3.14

https://youtu.be/qBXOgsZlCg4 ~savannahsolver

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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