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# Number N is randomly selected from a set of consecutive integers betwe

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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
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anairamitch1804 wrote:
Number N is randomly selected from a set of consecutive integers between 50 and 69, inclusive. What is the probability that N will have the same number of factors as 89?

a) 1/2
b 1/5
c) 0
d) 1/3
e) 1/4

The solution to this problem is straightforward if you know your prime numbers.
We need to determine the probability that a number when randomly selected from the set of integers between 50 and 69 inclusive will have the same number of factors as 89. Since 89 is prime (89 has only two factors), we need to determine the number of prime numbers between 50 and 69, inclusive. The prime numbers between 50 and 69, inclusive, are:

53, 57, 61, 67

We also know that there are 69 - 50 + 1 = 20 integers from 50 to 69, inclusive; thus, the probability that N will have the same number of factors as 89 is 4/20 = 1/5.

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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
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Hi All,

While this question becomes considerably easier IF you have your prime numbers memorized, even if you DON'T have them memorized, you can still get to the correct answer relatively quickly by doing some 'brute force' arithmetic and using the rules of division to avoid a lot of the implied 'math work.'

To start, we need to define how many integers are between 50 and 69, inclusive. This is an example of a 'fence post' problem - and if you recognize that, then you know that there are 69 - 50 + 1 = 20 total numbers. If you don't recognize that though, you can still 'group' the numbers....

50 - 59 = 10 numbers
60 - 69 = another 10 numbers
Total = 20 numbers

Next, we need to figure out the factors of 89. Since 89 is less than 100, the square-root of 89 is less than 10, so we just have to consider single-digit divisors... Since 89 is ODD, it's not divisible by any EVEN integers. Since it's digits add up to 17, it's not divisible by 3 or 9. Since it ends in a 9, it's not divisible by 5. All that's left is to try 7... which does NOT divide either. Thus, the only factors of 89 are 1 and 89.

After that, we have to determine how many of the 20 numbers have JUST TWO factors.

We can quickly eliminate all of the even numbers (they'll all have at least 4 factors).
We can also eliminate all of the multiples of 5 (they'll also have at least 4 factors).

So we're left with... 51, 53, 57, 59, 61, 63, 67 and 69

We can eliminate 51, 57, 63 and 69 with the 'rule of 3'

Now we're left with 53, 59, 61 and 67

This is 4 of the 20 numbers = 1/5 of the numbers. Based on the answer choices, the correct answer must be either B or C. Once you prove that ANY of those 4 numbers has just 2 factors, then you know what the correct answer must be...

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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
anairamitch1804 wrote:
Number N is randomly selected from a set of consecutive integers between 50 and 69, inclusive. What is the probability that N will have the same number of factors as 89?

a) 1/2
b 1/5
c) 0
d) 1/3
e) 1/4

total integers from 50 to 69 ; 69-50+1 = 20
prime integers =50,59,61 & 67
P = 4/20 ; 1/5
IMO B
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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
Isn't it 19, not numbers numbers? It says between 50 and 69, not including 50 and 69! So therefore it should not be 69-50+1, but in fact should be 68-51+1=18
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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
Hi ahamburger93,

The prompt tells us to consider the consecutive integers from 50 to 69, INCLUSIVE. This means that we have to include both 50 and 69, so there are 69 - 50 + 1 = 20 possible integers.

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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
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Re: Number N is randomly selected from a set of consecutive integers betwe [#permalink]
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