Difference between revisions of "1990 AHSME Problems/Problem 11"

 
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== 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 ==

Latest revision as of 04:26, 4 February 2016

Problem

How many positive integers less than $50$ have an odd number of positive integer divisors?

$\text{(A) } 3\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 11$

Solution

Divisors come in pairs, unless there is an integer square root, so we just need the perfect squares below $50$. There are $7$, so $\fbox{C}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
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

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