Difference between revisions of "1995 AHSME Problems/Problem 29"
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==Problem== | ==Problem== | ||
− | For how many three-element sets of positive integers <math>\{a,b,c\}</math> is it true that <math>a \times b \times c = 2310</math>? | + | For how many three-element sets of distinct positive integers <math>\{a,b,c\}</math> is it true that <math>a \times b \times c = 2310</math>? |
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==Solution 1== | ==Solution 1== | ||
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− | + | <math>2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11</math>. 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. | |
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+ | 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> | |
==Solution 2== | ==Solution 2== | ||
+ | The prime factorization of <math>2310</math> is <math>2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.</math> Therefore, we have the equation <cmath> abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,</cmath>where <math>a, b, c</math> must be distinct positive integers and order does not matter. There are <math>3</math> ways to assign each prime number on the right-hand side to one of the variables <math>a, b, c,</math> which gives <math>3^5 = 243</math> solutions for <math>(a, b, c).</math> However, three of these solutions have two <math>1</math>s and one <math>2310,</math> which contradicts the fact that <math>a, b, c</math> must be distinct. Because each prime factor appears only once, all other solutions have <math>a, b, c</math> distinct. Correcting for this, we get <math>243 - 3 = 240</math> ordered triples <math>(a, b, c)</math> where <math>a, b, c</math> are all distinct. | ||
− | <math> | + | Finally, since order does not matter, we must divide by <math>3!,</math> the number of ways to order <math>a, b, c.</math> This gives the final answer, <cmath>\frac{240}{3!} = \frac{240}{6} = \boxed{40}.</cmath> |
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==See also== | ==See also== | ||
{{AHSME box|year=1995|num-b=28|num-a=30}} | {{AHSME box|year=1995|num-b=28|num-a=30}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:33, 15 January 2023
Contents
Problem
For how many three-element sets of distinct positive integers is it true that
?
Solution 1
. 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 2
The prime factorization of is
Therefore, we have the equation
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,
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 |
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