Difference between revisions of "2005 AIME II Problems/Problem 4"
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Adding up the two cases, there are <math>44+391=\boxed{435}</math> positive integers. | Adding up the two cases, there are <math>44+391=\boxed{435}</math> positive integers. | ||
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== See also == | == See also == | ||
Latest revision as of 17:36, 1 January 2024
Contents
Problem
Find the number of positive integers that are divisors of at least one of 
Solution 1
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 
.
Solution 2
We can rewrite the three numbers as , , and .
Assume that 
(a positive integer) is a divisor of one of the numbers. Therefore, can be expressed as 
or as 
where 
, 
are in 
and 
, 
are positive integers.
If is the power of a single prime, then there are 11 possibilities (
to 
) for 
, 22 possibilities (
to 
) for 
, 10 possibilities (
to 
) for 
, and 1 possibility if 
. From this case, there are 
possibilities.
If is the product of the powers of two primes, then we can just multiply the exponents of each rewritten product to get the number of possibilities, since each exponent of the product must be greater than 0. From this case, there are 
possibilities.
Adding up the two cases, there are 
positive integers.
-alpha_2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
