1990 AHSME Problems/Problem 11: Difference between revisions
No edit summary |
No edit summary |
||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
<math>\fbox{C}</math> | Divisors come in pairs, unless there is an integer square root, so we just need the perfect squares below <math>50</math>. There are <math>7</math>, so <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
Revision as of 04:26, 4 February 2016
Problem
How many positive integers less than 
have an odd number of positive integer divisors?
Solution
Divisors come in pairs, unless there is an integer square root, so we just need the perfect squares below . There are 
, so 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
