Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
 
(One intermediate revision by the same user not shown)
Line 16: Line 16:
  
 
<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>.
 
<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>.
 +
<i>-bronzetruck2016<i>
  
 
==See Also==
 
==See Also==

Latest revision as of 19:17, 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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&oldid=151906"