Difference between revisions of "2005 AIME II Problems/Problem 4"

m (See Also)
m (Boxed the final answer.)
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Find the number of positive integers that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>
+
Find the number of [[positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>
  
 
== Solution ==
 
== Solution ==
Line 17: Line 17:
 
<math>\gcd(18^{11}, 10^{10}) = 2^{10} </math> which has 11 divisors.
 
<math>\gcd(18^{11}, 10^{10}) = 2^{10} </math> which has 11 divisors.
  
So now we have <math>121 + 64 + 276 - 8 -8 -11</math> potential divisors.  However, we've now undercounted those factors which divide all three of our numbers.  Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of <math>121 + 64 + 276 - 8 - 8 - 11 + 1 = 435</math>.
+
So now we have <math>121 + 64 + 276 - 8 -8 -11</math> potential divisors.  However, we've now undercounted those factors which divide all three of our numbers.  Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of <math>121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}</math>.
 
 
== See Also ==
 
 
 
*[[2005 AIME II Problems/Problem 3| Previous problem]]
 
*[[2005 AIME II Problems/Problem 5| Next problem]]
 
*[[2005 AIME II Problems]]
 
  
 +
== See also ==
 +
{{AIME  box|year=2005|n=II|num-b=3|num-a=5}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 01:17, 30 July 2015

Problem

Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$

Solution

$10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors.

$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.

$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.

Now, we use the Principle of Inclusion-Exclusion. We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise greatest common divisors.

$\gcd(10^{10},15^7) = 5^7$ which has 8 divisors.

$\gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors.

$\gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.

So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}$.

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png