Difference between revisions of "1990 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
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> | 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> | ||
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| + | == Solution 2 == | ||
== See also == | == See also == | ||
Revision as of 16:26, 1 January 2022
Contents
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 
Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
