Difference between revisions of "2005 AIME II Problems/Problem 4"
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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 == |
Latest revision as of 01:17, 30 July 2015
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 .
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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