Difference between revisions of "1995 AHSME Problems/Problem 29"
(→Solution 1) |
Potomaclion (talk | contribs) m (→Solution 2) |
||
| Line 14: | Line 14: | ||
We can account for permutations by assuming WLOG that <math>a</math> contains the prime factor 2. Thus, there are <math>3^4</math> ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to <math>a</math>, we have over counted each case twice, as for when we put certain prime factors into <math>b</math> and the rest into <math>c</math>, we count the exact same case when we put those prime factors which were in <math>b</math> into <math>c</math>. | We can account for permutations by assuming WLOG that <math>a</math> contains the prime factor 2. Thus, there are <math>3^4</math> ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to <math>a</math>, we have over counted each case twice, as for when we put certain prime factors into <math>b</math> and the rest into <math>c</math>, we count the exact same case when we put those prime factors which were in <math>b</math> into <math>c</math>. | ||
| − | Thus, our total number of cases is <math>\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}</math> | + | Thus, our total number of cases is <math>\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}.</math> |
==Solution 3== | ==Solution 3== | ||
Revision as of 15:53, 22 August 2021
Problem
For how many three-element sets of distinct positive integers 
is it true that 
?
Solution 1
Solution 2
. We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that 
contains the prime factor 2. Thus, there are 
ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to , we have over counted each case twice, as for when we put certain prime factors into 
and the rest into 
, we count the exact same case when we put those prime factors which were in into .
Thus, our total number of cases is 
Solution 3
The prime factorization of 
is 
Therefore, we have the equation ![\[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\]](http://latex.artofproblemsolving.com/e/0/8/e086833bfe720afd7048422a7dcb0742b323e206.png)
where 
must be distinct positive integers and order does not matter. There are 
ways to assign each prime number on the right-hand side to one of the variables 
which gives 
solutions for 
However, three of these solutions have two 
s and one 
which contradicts the fact that must be distinct. Because each prime factor appears only once, all other solutions have distinct. Correcting for this, we get 
ordered triples 
where are all distinct.
Finally, since order does not matter, we must divide by 
the number of ways to order 
This gives the final answer, ![\[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]](http://latex.artofproblemsolving.com/0/1/0/01036b8f3eee7df3d5b236a226f26705e62c1d2c.png)
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
