Difference between revisions of "2007 iTest Problems/Problem 34"
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Rockmanex3 (talk | contribs) (Solution to Problem 34) |
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Let <math>a/b</math> be the probability that a randomly selected divisor of <math>2007</math> is a multiple of <math>3</math>. If <math>a</math> and <math>b</math> are relatively prime positive integers, find <math>a+b</math>. | Let <math>a/b</math> be the probability that a randomly selected divisor of <math>2007</math> is a multiple of <math>3</math>. If <math>a</math> and <math>b</math> are relatively prime positive integers, find <math>a+b</math>. | ||
| − | == Solution == | + | ==Solution== |
| + | The prime factorization of <math>2007</math> is <math>3^2 \cdot 223</math>, so <math>2007</math> has <math>6</math> positive divisors. Of the six positive divisors, four of them are divisible by <math>3</math>. Thus, the probability that a divisor is a multiple of <math>3</math> is <math>\frac{4}{6} = \frac{2}{3}</math>, so <math>a + b = \boxed{5}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{iTest box|year=2007|num-b=33|num-a=35}} | ||
| + | |||
| + | [[Category:Introductory Probability Problems]] | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 21:24, 10 June 2018
Problem
Let 
be the probability that a randomly selected divisor of 
is a multiple of 
. If 
and 
are relatively prime positive integers, find 
.
Solution
The prime factorization of is 
, so has 
positive divisors. Of the six positive divisors, four of them are divisible by . Thus, the probability that a divisor is a multiple of is 
, so 
.
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 33 |
Followed by: Problem 35 | |
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