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

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== Solution ==
 
== Solution ==
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<math>10^{10} = 2^{10}\cdot 5^{10}</math> so <math>10^{10}</math> has <math>11\cdot11 = 121</math> [[divisor]]s.
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<math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisors.
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<math>18^{11} = 2^{11}\cdot3^{22}</math> so <math>18^{11}</math> has <math>12\cdot23 = 276</math> divisors.
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Now, we use the [[Principle of Inclusion-Exclusion]].  We have <math>121 + 64 + 276</math> 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 divisor]]s.
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<math>\gcd(10^{10},15^7) = 5^7 </math> which has 8 divisors.
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<math>\gcd(15^7, 18^{11}) = 3^7 </math> which has 8 divisors.
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<math>\gcd(18^{11}, 10^{10}) = 2^{10} </math> which has 11 divisors.
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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>.
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== See Also ==
 
== See Also ==
 
*[[2005 AIME II Problems]]
 
*[[2005 AIME II Problems]]

Revision as of 10:02, 21 July 2006

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 $121 + 64 + 276 - 8 - 8 - 11 + 1 = 435$.

See Also

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