Difference between revisions of "2012 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | 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. So we are done, and the answer is <math> \boxed{\textbf{(A)}\ 3127}</math>. | + | 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 == | ||
| + | https://youtu.be/HISL2-N5NVg?t=526 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:58, 10 November 2023
Contents
Problem
What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
Solution
The problem states that the answer cannot be a perfect square or have prime factors less than 
. Therefore, the answer will be the product of at least two different primes greater than . The two smallest primes greater than are 
and 
. Multiplying these two primes, we obtain the number 
, which is also the smallest number on the list of answer choices.
So we are done, and the answer is 
.
Video Solutions
https://youtu.be/HISL2-N5NVg?t=526
~ pi_is_3.14
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. 
