Difference between revisions of "2008 iTest Problems/Problem 29"
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==Solution 2== | ==Solution 2== | ||
+ | <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>. | ||
==See Also== | ==See Also== |
Revision as of 20:15, 18 April 2021
Contents
Problem
Find the number of ordered triplets of positive integers such that (the product of , and is ).
Solution 1
The number can be factored into . Use casework to organize the counting.
- If two numbers are , then the third one must be , and there are ways to write the ordered pairs.
- If one number is a , then there are possible pairs of numbers for the other two. Since the numbers are all different, there are ways to write the ordered pairs.
- If none of the numbers are , 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 and , and there are ways in this case.
Altogether, there are ordered pairs that satisfy the criteria.
Solution 2
can be prime factorized into . We can think of each ordered pair 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 ways to assign three 2s to three distinct letters, and there are ways to assign one 251 to three distinct letters. Multiplying, we get .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 28 |
Followed by: Problem 30 | |
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