Difference between revisions of "1995 AHSME Problems/Problem 29"
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<math> \mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 } </math> | <math> \mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 } </math> | ||
| − | ==Solution== | + | ==Solution 1== |
<math>2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11</math>. The number of ordered triples <math>(x,y,z)</math> with <math>xyz = 2310</math> is therefore <math>3^5</math>, since each prime dividing 2310 divides exactly one of <math>x,y,z</math>. | <math>2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11</math>. The number of ordered triples <math>(x,y,z)</math> with <math>xyz = 2310</math> is therefore <math>3^5</math>, since each prime dividing 2310 divides exactly one of <math>x,y,z</math>. | ||
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The number of sets of distinct integers <math>\{ a,b,c\}</math> such that <math>abc = 2310</math> is therefore <math>\frac {3^5 - 3}{6}</math> (accounting for rearrangement), or <math>\boxed{40}</math>. | The number of sets of distinct integers <math>\{ a,b,c\}</math> such that <math>abc = 2310</math> is therefore <math>\frac {3^5 - 3}{6}</math> (accounting for rearrangement), or <math>\boxed{40}</math>. | ||
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| + | ==Solution 2== | ||
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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>. | ||
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| + | Thus, our total number of cases is <math>\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}</math> | ||
==See also== | ==See also== | ||
{{AHSME box|year=1995|num-b=28|num-a=30}} | {{AHSME box|year=1995|num-b=28|num-a=30}} | ||
Revision as of 16:10, 7 February 2009
Contents
Problem
For how many three-element sets of positive integers 
is it true that 
?
Solution 1
. The number of ordered triples 
with 
is therefore 
, since each prime dividing 2310 divides exactly one of 
.
Three of these triples have two of equal (namely when one is 2310 and the other two are 1). So there are 
with distinct.
The number of sets of distinct integers 
such that 
is therefore 
(accounting for rearrangement), or 
.
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 
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 | ||
