Number N is randomly selected from a set of consecutive inte

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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

Pleas help with this problem.

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by DavidG@VeritasPrep » Sun Jan 08, 2017 4:47 pm
Anaira Mitch 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

Pleas help with this problem.
Total: We know there are 20 integers between 50 and 69 inclusive, so that's our total.

Desired: 89 has 2 factors, 1 and itself. So we're looking for all the numbers in this range that have 2 factors. Put another way, we're looking for the number of primes. Let's list them out. 53, 59, 61, 67. There are 4 prime integers between 50 and 69.

4/20 = 1/5. The answer is B
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by Jay@ManhattanReview » Sun Jan 08, 2017 9:39 pm
Anaira Mitch 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

Pleas help with this problem.
We see that 89 is a prime number and thus have only two factors: 1 and 89.

Each prime number has two factors: 1 and the number itself. Thus, ONLY the prime numbers between 50 and 69, inclusive, will also have only two factors.

Let us understand how to be sure that whether a number is prime or not.

Step 1: Take a number. Say, N = 53.
Step 2: Take the root of the number, thus n = square root of N = √N. n = √53 = ~7...
Step 3: Note the prime numbers less than n. n = 7, thus prime numbers < 7... are: 2, 3, 5, and 7.
Step 4: Divide the number, N by each of the prime numbers noted in Step 3. If the number, N is divisible by any of the prime numbers, the number is non-prime, else a prime. Since n = 53 is not divisible by 2, 3, 5, or 7, 53 is prime.

There are four prime numbers between 50 and 69 inclusive: 53, 59, 61, and 67, and there are 20 integers between 50 and 69 inclusive.

Thus, the probability of choosing a prime number out of 20 numbers = 4/20 = 1/5.

Answer: B.

Hope this helps!

-Jay
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by Jeff@TargetTestPrep » Wed Jan 11, 2017 7:10 am
Anaira Mitch 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.

Answer: B

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by [email protected] » Wed Jan 11, 2017 10:55 am
Hi Anaira Mitch,

As several of the other explanations have pointed it, this question becomes considerably easier IF you have your prime numbers memorized. However, 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 the square-root of 89 is less than 10, 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...

Final Answer: B

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by Matt@VeritasPrep » Thu Jan 19, 2017 9:24 pm
Once you know that you're looking for primes, you're left with the problem of determining whether some numbers in a given range are prime.

A nice trick for eliminating the bulk of them is to look for multiples of possible prime factors of those numbers. When considering numbers from 50 to 69, we only need to consider the possible prime factors 2, 3, 5, and 7. Any even number will divide by 2 and anything that ends in 5 will divide by 5, so we're left with:

51, 53, 57, 59, 61, 63, 67, 69

We start by eliminating multiples of 3, leaving

53, 59, 61, 67

Then we look for multiples of 7, of which are there are none remaining. So we've got FOUR primes in range, out of the 20 numbers in the set, giving us 4/20, or 1/5.

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by Matt@VeritasPrep » Thu Jan 19, 2017 9:27 pm
You might also ask why I didn't consider any larger primes, such as 11, 13, ...

When testing whether a number is prime, start by thinking about its possible factors. A friendly number like 64 helps illustrate. If I think of 64 as the product of two of its factors, I get a number of possibilities:

1 * 64
2 * 32
4 * 16
8 * 8

Notice how ONE of the two numbers in the pair is always less than or equal to the square root of 64? This will be true of any integer: if it has other factors between 1 and itself, at least one of those factors must be less than or equal to the number's square root.

Since the largest term in our set is 69, we can use √69, or ≈ 8 as a guide. If a positive integer less than 69 has some other factors, one of those factors MUST be less than ≈ 8. So we only need to consider 2, 3, 5, and 7.