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

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Consider divisors of <math>n^2: a,b</math> such that  
 
Consider divisors of <math>n^2: a,b</math> such that  
 
<math>ab=n^2</math>.
 
<math>ab=n^2</math>.
WLOG, let <math>b\ge{a}. b=\frac{n}{a}
+
WLOG, let <math>b\ge{a}. b=\frac{n}{a}</math>
  
Then, it is easy to see that </math>a<math> will always be less than </math>b<math> as we go down the divisor list of </math>n^2<math> until we hit </math>n<math>.  
+
Then, it is easy to see that <math>a</math> will always be less than <math>b</math> as we go down the divisor list of <math>n^2</math> until we hit <math>n</math>.  
  
Therefore, the median divisor of </math>n^2<math> is </math>n<math>.
+
Therefore, the median divisor of <math>n^2</math> is <math>n</math>.
  
Then, there are </math>(63)(39)=2457<math> divisors of </math>n^2<math>. Exactly </math>\frac{2457-1}{2}=1228<math> of these divisors are </math><n<math>
+
Then, there are <math>(63)(39)=2457</math> divisors of <math>n^2</math>. Exactly <math>\frac{2457-1}{2}=1228</math> of these divisors are <math><n</math>
  
There are </math>(32)(20)-1=639<math> divisors of </math>n<math> that are </math><n<math>.
+
There are <math>(32)(20)-1=639</math> divisors of <math>n</math> that are <math><n</math>.
  
Therefore, the answer is </math>1228-639=\boxed{589}$.
+
Therefore, the answer is <math>1228-639=\boxed{589}</math>.
  
  

Revision as of 00:46, 20 August 2018

Problem

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

Solution 1

We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$. There are $32\times20-1 = 639$ factors of $n$, which clearly are less than $n$, but are still factors of $n$. Therefore, using complementary counting, there are $1228-639=\boxed{589}$ factors of $n^2$ that do not divide $n$.

Solution 2

Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$. Then $n^2$ has $\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$.

This simplifies to $\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$.

The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$.

Thus, our general formula for $n=p_1^{k_1}p_2^{k_2}$ is

Number of factors that satisfy the above $=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2$

Incorporating this into our problem gives $19\times31=\boxed{589}$.

Solution 3

Consider divisors of $n^2: a,b$ such that $ab=n^2$. WLOG, let $b\ge{a}. b=\frac{n}{a}$

Then, it is easy to see that $a$ will always be less than $b$ as we go down the divisor list of $n^2$ until we hit $n$.

Therefore, the median divisor of $n^2$ is $n$.

Then, there are $(63)(39)=2457$ divisors of $n^2$. Exactly $\frac{2457-1}{2}=1228$ of these divisors are $<n$

There are $(32)(20)-1=639$ divisors of $n$ that are $<n$.

Therefore, the answer is $1228-639=\boxed{589}$.


See also

1995 AIME (ProblemsAnswer KeyResources)
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

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