How many prime numbers between 1 and 100 are factors of 7,15

How many prime numbers between 1 and 100 are factors of 7,150?
A. One
B. Two
C. Three
D. Four
E. Five

Make prime factorization of 7,150 --> 7,150=2*5^2*11*13. So 4 prime numbers between 1 and 100 (namely 2, 5, 11, and 13) are factors of 7,150 (you shouldn't count one prime factor twice).

Answer: D.

Hate to play devil's advocate here, but the question doesn't clarify how many UNIQUE prime numbers are factors of 7150. (Which would change the answer to five). Am I missing something?

Yes, you are! Read the question again:
How many prime numbers between 1 and 100 are factors of 7,150?

The prime numbers between 1 and 100 are 2, 3, 5, 7, 11, 13, 17, 19... etc

5 appears only once between 1 and 100 so there is absolutely no confusion. Out of these 25 prime numbers, only 4 are factors of 7150: 2, 5, 11 and 13

I also got 2,5,7,11 and 13 but it took me more than the two minutes to break down 143 into 11 and 13. 2,5 and 7 were the easier ones. I'm sure many will agree at in times of stress, the mind starts to play tricks. At first I stopped after 7 and thought that 143 cannot be factored further and chose C as the answer i.e. 3 prime factors but then got the other two in over the two mins.

Therefore, is there a quick way to get the prime factors of a large number?

Thanks.

So, just to clarify this, if the question only asks for the factors we count all of the factors, no matter if there are repetitions.

Only when it specifically says that we need the "unique" factors should we diregard repeating factors.

Right?

Not really.
Take the example of factors of 8:

How many total factors does 8 have and what are they?
They are 1, 2, 4 and 8 - a total of 4 factors

We know that 8 = 2^3 but we don't say that factors are 8 are 1, 2, 2, 2, 4 and 8.

Similarly, if we are asked - how many prime factors does 8 have? I will answer only 1 (the prime factor is 2). The number of prime factors of 8 are not 3 (not 2, 2, 2). I know of people who are not very convinced with this and hence, I assume that GMAT will insert the word "unique" to remove all doubts.

On GMAT, I would expect it to be - How many unique prime factors does 8 have?

In the original question, there is no doubt since they ask "how many prime numbers are factors of..." The set of prime numbers does not have multiple entries and hence there is no doubt that we are talking about unique prime factors only.

Bunuel
Bunuel,

Do you have any fast tricks to understand that 7,150 is divisible by 11 and 13?

The only silly mistake I made was to miss 11 and 13 as prime factors of 7150.

Thank you!

Hi elisabettaportioli,

The speed with which you 'prime-factor' 7150 into its 'pieces' is likely going to be influenced by the 'first' number you factor out.

Looking at 7150, you could easily start with a 2 (because 7150 is even), a 5 (because 7150 ends in a 0) or a 10 (also since it ends in a 0).

I actually started with 50, since 50 divides into 100 twice.....7100 = (71x2) fifties....

So 7150 = 142 fifties + 1 fifty =
(50)(143)

The (50) can be quickly broken down into (2)(5)(5)

Now, looking at the 143, we know that NO even numbers will divide in (since even numbers do NOT divide into odd numbers). If you know the 'rule of 3', then you know that 3 does NOT divide in. Since 3 doesn't divide in, 9 won't divide in either. 5 won't divide in for obvious reasons. Thus, we're really left with just a handful of possibilities:

1) 143 might be prime
2) 7, 11 and/or 13 might divide in

It's pretty easy to eliminate 7 as an option (it divides into 14, but not 3). Once you find that 11 divides in, you end up with the 13 by default.

GMAT assassins aren't born, they're made,
Rich

EMPOWERgmatRichC, Bunuel, VeritasPrepKarishma:
I stopped at 143 as felt it was a prime number after trying 3, 5, 7 n 9. I didnt go to 11. If i had taken that one extra step I would have gotten it right. But how far one should go? Here it is 11 but on some other questions it could be 17, 19, 23 etc. Is there a neat formula too tell all prime factors of a number? thanks

Hi NoHalfMeasures,

When you say that you tried 3, 5, 7 and 9, what actual 'work' did you do? I ask because your Quant skills are clearly strong (you have a Q48 to prove it), so I would guess that you would have figured out rather quickly that those single-digit primes wouldn't divide into 143.

Since those smaller primes don't divide in, there can't be that many left to check out (and checking EITHER 11 or 13 would have been enough work to correctly answer the question). At it's core, this is a test of 'thoroughness' - it's true that most people wouldn't think to try dividing 11, but most people can't score Q48+, so you have to decide what 'extra work' (if any) YOU'RE willing to do to be thorough, prove that your answer is correct and pick up those extra points.

GMAT assassins aren't born, they're made,
Rich

EMPOWERgmatRichC, Bunuel, VeritasPrepKarishma:

Here is my method (and I am guessing this is not the right approach, but wanted to confirm):

7150 = 7000 + 150
7000 = 7 * 1000 (so we have 7, 5 & 2)
150 = 3 * 50 (so we have 3, 5 & 2)

combining both we have 2, 3, 5 & 7 as prime numbers.

At the end, I am still getting the same number of prime numbers for 7150, albeit different values (maybe coincidence).

I will be trying this with the next similar questions I come across, till then any help....greatly appreciated.

Hi colorblind,

Unfortunately, that logic is NOT correct (and it was lucky that you ended up with the correct answer). Here's a simple example that proves the logic is NOT correct.

17 has one prime factor: 17

Using the logic you described...
17 = 10+7
10 has prime factors of 2 and 5 and 7 has one prime factor: 7

However, these prime factors (re: 2, 5 and 7) are NOT the same prime factors of 17 (re: 17) and the NUMBER of prime factors is also different.

GMAT assassins aren't born, they're made,
Rich

We start by prime factoring 7,150.

7,150 = 715 x 10 = 143 x 5 x 10

At this point we must be careful. They WANT you to think that 143 is prime. Using our divisibility rules, however, we can determine that 143 is divisible by 11. A number is divisible by 11 if the sum of the odd-numbered place digits minus the sum of the even-numbered place digits is divisible by 11. We can test 143 to prove this:

1 + 3 – 4 = 4 – 4 = 0

Since zero is divisible by 11, we know that 143 is divisible by 11. We can now finish the prime factorization.

143 x 5 x 10 = 11 x 13 x 5 x 5 x 2

11 x 13 x 5^2 x 2

Thus we can see that there are 4 different prime factors in 7,150.

Answer is D.

Using prime factorization we get: 7,150 = (2)(5)(5)(11)(13)

So, FOUR prime numbers (2, 5, 11 and 13) are factors of 7,150

Answer: D

Cheers,
Brent

ASIDE: For anyone who feels that the correct answer should be E (since there are two 5's in the prime factorization of 7150), here's why the correct answer is actually D

The question starts with "How many prime numbers between 1 and 100..."
The prime numbers between 1 and 100 are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

"....are factors of 7,150"

So, how many numbers from the set {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97} are factors of 7150?

The prime factors of 7150 are underlined {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97} are factors of 7150

There are FOUR underlined primes

My answer was 5 because I thought 1 is included !!!

Hi Elyazid,

As a general rule, when the answers are consecutive integers, you should be a bit more careful about your work (and confirming that you are answering the question that is ASKED) - since if you are 'off' even a little, then you will select one of the wrong answers and not even realize it. Here, the specific question asks for the number of PRIME numbers (and '1' is NOT a prime). It's fine to make this mistake now - and the 'takeaway' here for your future studies is to always make sure to write the specific question on your note-pad, so that you are always thinking in terms of answering THAT question.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

We need to find how many prime numbers between 1 and 100 are factors of 7,150

If you want to know what is a prime number then watch this video

Now, we can factorize 7150 to get

7150 = 715*10 = 5*143*5*2 = 5*11*13*5*2 = 2*\(5^2\)*11*13

Clearly, there are 4 prime numbers 2, 5, 11 and 13 which are between 1 and 100 and are factors of 7,150

So, Answer will be D.
Hope it helps!

Watch the following video to learn Basics of Prime Numbers




Watch the following video to learn How to find Prime Numbers from 1 to 100