Difference between revisions of "1984 AHSME Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | Let <math> n </math> be the smallest nonprime integer greater than <math> 1 </math> with no prime factor less than <math> 10 </math>. Then | + | Let <math> n </math> be the smallest nonprime [[integer]] greater than <math> 1 </math> with no [[Prime factorization|prime factor]] less than <math> 10 </math>. Then |
<math> \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 </math> | <math> \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 </math> | ||
==Solution== | ==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>. | + | 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>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=2|num-a=4}} | {{AHSME box|year=1984|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:46, 8 November 2016
Problem
Let 
be the smallest nonprime integer greater than 
with no prime factor less than 
. Then
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 . These prime factors must be greater than , so the least prime factor is 
. Therefore, the least integer is 
, which is in 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
