Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 AIME Problems/Problem 6"

(→‎Solution)
m (fbox not rendering for me, + box, grammar)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>\displaystyle n=2^{31}3^{19}.</math>  How many positive integer divisors of <math>\displaystyle n^2</math> are less than <math>\displaystyle n_{}</math> but do not divide <math>\displaystyle n_{}</math>?
+
Let <math>\displaystyle n=2^{31}3^{19}.</math>  How many positive [[integer]] [[divisor]]s of <math>\displaystyle n^2</math> are less than <math>\displaystyle n_{}</math> but do not divide <math>\displaystyle n_{}</math>?
  
 
== Solution ==
 
== Solution ==
−
We know that <math>n^2</math> must have <math>63\times 39</math> factors by its prime factorization. There are <math>\frac{63\times 39-1}{2} = 1228</math> factors of <math>n^2</math> that are less than <math>n</math>, because if they form pairs <math>a</math>, then there is one factor per pair that is less than <math>n</math>. There are <math>32\times20-1 = 639</math> factors of <math>n</math> that are less than <math>n</math> itself. These are also factors of <math>n^2</math>. Therefore, there are <math>1228-639=\fbox{539}</math> factors of <math>n</math> that does not divide <math>n</math>.
+
We know that <math>n^2</math> must have <math>63\times 39</math> [[factor]]s by its [[prime factorization]]. There are <math>\frac{63\times 39-1}{2} = 1228</math> factors of <math>n^2</math> that are less than <math>n</math>, because if they form pairs <math>a</math>, then there is one factor per pair that is less than <math>n</math>. There are <math>32\times20-1 = 639</math> factors of <math>n</math> that are less than <math>n</math> itself. These are also factors of <math>n^2</math>. Therefore, there are <math>1228-639=539</math> factors of <math>n</math> that do not divide <math>n</math>.
  
 
== See also ==
 
== See also ==
−
* [[1995 AIME Problems/Problem 5 | Previous problem]]
 
−
* [[1995 AIME Problems/Problem 7 | Next problem]]
 
 
* [[1995 AIME Problems]]
 
* [[1995 AIME Problems]]
 +
 +
{{AIME box|year=1995|num-b=5|num-a=7}}

Revision as of 19:35, 8 February 2007

Problem

Let $\displaystyle n=2^{31}3^{19}.$ How many positive integer divisors of $\displaystyle n^2$ are less than $\displaystyle n_{}$ but do not divide $\displaystyle n_{}$?

Solution

We know that $n^2$ must have $63\times 39$ factors by its prime factorization. There are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$, because if they form pairs $a$, then there is one factor per pair that is less than $n$. There are $32\times20-1 = 639$ factors of $n$ that are less than $n$ itself. These are also factors of $n^2$. Therefore, there are $1228-639=539$ factors of $n$ that do not divide $n$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_6&oldid=12996"