Difference between revisions of "2008 iTest Problems/Problem 29"

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Altogether, there are <math>\boxed{30}</math> ordered pairs that satisfy the criteria.
 
Altogether, there are <math>\boxed{30}</math> ordered pairs that satisfy the criteria.
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==Solution 2==
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<math>2008</math> can be prime factorized into <math>2^3\cdot251</math>. We can think of each ordered pair <math>(a,b,c)</math> as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a "assign non-distinct objects to distinct piles" problem. In problems like this, we should use [[stars and bars]]. There are <math>\binom{5}{2}=10</math> ways to assign three 2s to three distinct letters, and there are <math>\binom{3}{2}=3</math> ways to assign one 251 to three distinct letters. Multiplying, we get <math>3\cdot10=\boxed{30}</math>.
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<i>-bronzetruck2016<i>
  
 
==See Also==
 
==See Also==

Latest revision as of 20:17, 18 April 2021

Problem

Find the number of ordered triplets $(a,b,c)$ of positive integers such that $abc=2008$ (the product of $a, b$, and $c$ is $2008$).

Solution 1

The number $2008$ can be factored into $2^3 \cdot 251$. Use casework to organize the counting.

  • If two numbers are $1$, then the third one must be $2008$, and there are $3$ ways to write the ordered pairs.
  • If one number is a $1$, then there are $\tfrac{8-2}{2} = 3$ possible pairs of numbers for the other two. Since the numbers are all different, there are $3 \cdot 6 = 18$ ways to write the ordered pairs.
  • If none of the numbers are $1$, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are $2,2,502$ and $2,4,251$, and there are $3+6=9$ ways in this case.

Altogether, there are $\boxed{30}$ ordered pairs that satisfy the criteria.

Solution 2

$2008$ can be prime factorized into $2^3\cdot251$. We can think of each ordered pair $(a,b,c)$ as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a "assign non-distinct objects to distinct piles" problem. In problems like this, we should use stars and bars. There are $\binom{5}{2}=10$ ways to assign three 2s to three distinct letters, and there are $\binom{3}{2}=3$ ways to assign one 251 to three distinct letters. Multiplying, we get $3\cdot10=\boxed{30}$. -bronzetruck2016

See Also

2008 iTest (Problems)
Preceded by:
Problem 28
Followed by:
Problem 30
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