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Difference between revisions of "2005 AIME II Problems/Problem 4"
m (Boxed the final answer.)
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<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 ==
 
== See also ==

Revision as of 00:17, 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 (Problems • Answer Key • Resources)
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

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