Difference between revisions of "2005 AIME II Problems/Problem 4"
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== Solution == | == Solution == | ||
+ | <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
Solution
so
has
divisors.
so
has
divisors.
so
has
divisors.
Now, we use the Principle of Inclusion-Exclusion. We have 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.
which has 8 divisors.
which has 8 divisors.
which has 11 divisors.
So now we have 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
.