Probability that a randomly chosen 3 digit number has exactly 3 factor

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Homework Statement



The probability that a randomly chosen 3 digit number has exactly 3 factors is

1)2/225
2)9/900
3)1/800
4)None of these


The Attempt at a Solution



No idea.
 

Discussion

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EDIT: Rethinking of the question, it's a bit tricky, depending on the way we interpret the question.

1/ If we count 1 and the number itself as its factors, then the only way to construct such number is A = B2 where B is a prime number. Then B should range from 10 to 31. Count the number of possible B's.

2/ If we exclude 1 and the number itself, then there is no way we can construct such number (Why?).

3/ If we exclude 1, then A = BC, where B and C are prime numbers (Why only 2 numbers? Why prime?). However, counting prime numbers up to 999 is a very very tough task :biggrin:
 
Last edited:


First find how many 3 digit numbers are there.
 


@jbunniii: 105 is divisible by 3, 5, 7 and 3*5, 3*7, 5*7, so how many factors are there? :wink:

P.S.: These are under the assumption that we don't count negative factors. Even if we do, it would be pointless, as there is no number with odd number of factors if negative factors are included.
 


hikaru1221 said:
@jbunniii: 105 is divisible by 3, 5, 7 and 3*5, 3*7, 5*7, so how many factors are there? :wink:

P.S.: These are under the assumption that we don't count negative factors. Even if we do, it would be pointless, as there is no number with odd number of factors if negative factors are included.

I assumed that the question meant prime factors, but you're right, it doesn't explicitly say that. Nor does it say whether the factors have to be distinct. Nor, as you pointed out, whether 1 and the number itself count, nor whether negative numbers are allowed.

Because of this ambiguity, "none of the above" is the most defensible answer.
 


Just a friendly reminder to y'all that this is a Homework Help forum, and we are not allowed to do the OP's work for him. Please stick to hints and questions, and ensure that the OP does the bulk of the work. Thanks.
 


Sorry, I did not recognise that it was a solution :). The thread looked closed as everybody agreed that it was not well-worded, so I just wanted to add a comment.

Next time I will be more careful...

ehild
 


hikaru1221 said:
EDIT: Rethinking of the question, it's a bit tricky, depending on the way we interpret the question.

1/ If we count 1 and the number itself as its factors, then the only way to construct such number is A = B2 where B is a prime number. Then B should range from 10 to 31. Count the number of possible B's.

2/ If we exclude 1 and the number itself, then there is no way we can construct such number (Why?).

3/ If we exclude 1, then A = BC, where B and C are prime numbers (Why only 2 numbers? Why prime?). However, counting prime numbers up to 999 is a very very tough task :biggrin:
This is a good analysis. Such things improve your math.

Stephen Tashi said:
Abdul,

Where you get these problems that you have posted recently?

All questions are either from various practice books available in my country or test questions which I wrongly answered/did not attempt. If you want such questions, google (books) for AIEEE Mathematics. If you like solving the toughest math of class 11th/12th, google for IIT-JEE Mathematics :smile:

berkeman said:
Just a friendly reminder to y'all that this is a Homework Help forum, and we are not allowed to do the OP's work for him. Please stick to hints and questions, and ensure that the OP does the bulk of the work. Thanks.

I guess ehild got a warning too :biggrin: