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Re: How many prime numbers between 1 and 100? [#permalink]

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16 Jan 2012, 12:09

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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?

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?

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).

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
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Re: How many prime numbers between 1 and 100 are factors of 7150 [#permalink]

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13 Feb 2012, 19:09

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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?

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.

Generally there is no easy way to check whether some very large number is a prime (well if it doesn't have some small primes, which are easy to check). You'll need a computer to do this.

Next, the GMAT won't give you a large number to factorize if there is no shortcut for that.

For example in our original question after you find 2, 5, and 7, just check for the next prime 11: 143/11=13.

You can also use divisibility rule for 11: if you sum every second digit and then subtract all other digits and the answer is divisible by 11, then the number is divisible by 11. So, for 143: (1+3)-4=0 --> 0 is divisible by 11 thus 143 is divisible by 11.

Or, you can notice that 143=130+13=13*10+13, so 143 must be divisible by 13.

So, as you can see there are plenty of shortcuts to get prime factorization of the numbers from the GMAT problems.

For more on divisibility rules and on verifying the primality check Number Theory chapter of Math Book: math-number-theory-88376.html

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.

Sadly, you will need to check for all prime numbers till the square root of the given number. We don't know the square root of 143 but we can approximate it. 143 is very close to 144. The square root of 144 is 12 so you will need to check for all prime numbers less than 12. If none of the prime numbers less than 12 is a factor of 143, then you can say that 143 is prime. So the question is not over till you don't check for 11 too.

Re: How many prime numbers between 1 and 100 are factors of 7150 [#permalink]

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28 Feb 2014, 06:43

I got the factorization in a much simpler way:

Multiples of 5 are extremely simple to factor; just multiply the number by 2 and divide by 10. In this case, I knew that 715 was a multiple of 5, so I knew 10*715 could be broken down into 10*(143*5). From there, I recognized 143 as a number that appears frequently on the gmat, and tested multiples of 11 to find 11 & 13. Then, the breakdown is simple: 2*5*5*11*13 = 4 distinct primes.

Answer: D

This took me 32 seconds to solve, so there are definitely simple methods to this type of problem.

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.
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