Difference between revisions of "1984 AHSME Problems/Problem 3"

Revision as of 19:30, 7 November 2016

Problem

Let $n$ be the smallest nonprime integer greater than $1$ with no prime factor less than $10$ . Then

$\mathrm{(A) \ }100<n\leq110 \qquad \mathrm{(B) \ }110<n\leq120 \qquad \mathrm{(C) \ } 120<n\leq130 \qquad \mathrm{(D) \ }130<n\leq140 \qquad \mathrm{(E) \ } 140<n\leq150$

Solution

To solve the problem, you would have to find the smallest prime number greater than ten: eleven. So, the smallest number with eleven as prime factorization and greater than 100 = 11^2 (i.e. 121). Which is in $\boxed{\text{C}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==Solution==
-Since the number isn't [[prime]], it is a product of two primes. If the least integer were a product of more than two primes, then one prime could be removed without making the number prime or introducing any prime factors less than <math> 10 </math>. These prime factors must be greater than <math> 10 </math>, so the least prime factor is <math> 11 </math>. Therefore, the least integer is <math> 11^2=121 </math>, which is in <math> \boxed{\text{C}} </math>.
+To solve the problem, you would have to find the smallest prime number greater than ten: eleven. So, the smallest number with eleven as prime factorization and greater than 100 = 11^2 (i.e. 121). Which is in <math> \boxed{\text{C}} </math>.
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Difference between revisions of "1984 AHSME Problems/Problem 3"

Revision as of 19:30, 7 November 2016

Problem

Solution

See Also