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How many prime numbers exist between 200 and 220?  [#permalink]

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Question Stats: 41% (02:09) correct 59% (02:06) wrong based on 234 sessions

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How many prime numbers exist between 200 and 220?

(A) None
(B) One
(C) Two
(D) Three
(E) Four
Director  B
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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2
2
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?
Math Expert V
Joined: 02 Sep 2009
Posts: 61385
Re: How many prime numbers exist between 200 and 220?  [#permalink]

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kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

I think you did everything right.
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav Director  B
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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1
1
GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav The rule is if X is the number, we need to check the prime factors till \sqrt{x}
Math Expert V
Joined: 02 Aug 2009
Posts: 8250
Re: How many prime numbers exist between 200 and 220?  [#permalink]

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1
GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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chetan2u wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..

Thanks and kudos to both of you.

Regards,
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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One more rul regarding prime no

All the prime no would be in the form of 6n +/- 1.

Cheers,
Rajan
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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1
rajsinghal71 wrote:
One more rul regarding prime no

All the prime no would be in the form of 6n +/- 1.

Cheers,
Rajan

Hi rajan,

It's good that you know some properties like you mentioned here that every prime Number Greater than 3 can be represented in form of 6k+1 or 6k-1 but since it's a GMAT forum so let me make it clear here that no such property is expected to be used by students in GMAT because GMAT checks very basic skills of mathematics and doesn't expect students to be loaded with various properties.

P.S. Also for the reader, Please Don't use this property as Final check of a Number being Prime, however it can be a primary check to suspect if the number is prime subjected to the number satisfying this property.
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav Check out these two posts:

http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/

The second one ends with an explanation of this concept. Will help you understand a lot of things about factors and their placement.
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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1
GauravSolanky wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..

Thanks and kudos to both of you.

Regards,
Gaurav The check for a Number to be prime is "If the number to be checked is NOT divisible by any prime number Less than the square root of the number then it is said to he a Prime Number"

I.e. 219 will be prime if 219 is NOT divisible by any prime number less than "Square root of 219"

Square root if 219 = approx.14.8

I.e. check if 219 is divisible by any one of 2,3,5,7,11,13. If 219 is not divisible by any one of these prime numbers which are less than 14.8 then 219 will be prime.
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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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VeritasPrepKarishma wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel

I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime

As 220 < 225, so maximum prime I need to check for divisibility is 13.

Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7

The last step took some time. Any idea if we can use anything else?

Can you please throw more light on the logic behind this ?

As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC

Thanks,
Gaurav Check out these two posts:

http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/

The second one ends with an explanation of this concept. Will help you understand a lot of things about factors and their placement.

Thanks Karishma, both articles were really helpful.

Regards,
Gaurav
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Joined: 24 Jun 2008
Posts: 2012
Re: How many prime numbers exist between 200 and 220?  [#permalink]

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GMATinsight wrote:

The check for a Number to be prime is "If the number to be checked is NOT divisible by any prime number Less than the square root of the number then it is said to he a Prime Number"

That should read "less than or equal to the square root". A number like 169 is not divisible by any prime less than √169, but 169 is still not a prime.
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GRE 1: Q169 V154 Re: How many prime numbers exist between 200 and 220?  [#permalink]

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1
Excellent Question.
Here is what i did on this one ->

In order to check whether a given number is prime or not => We must divide it by all the primes less than or equal to the square root of that number.
PROPERTY => IF N IS NOT DIVISIBLE BY ANY PRIMES LESS THAN OR EQUAL TO THE SQAURE ROOT OF N => N is PRIMES.

Here the highest number =220
$$√220$$=> 14.something

Primes upto 14.something => {2,3,5,7,11,13}

Next => ALL PRIMES >5 MUST HAVE {1,3,7,9} as its UNITS DIGIT.

So lets check =->>

201=> Divisible by 3 => Rejected
203=>Divisible by 7 => Rejected
207=>Divisible by 3 => Rejected
209=>Divisible by 11 => Rejected
211=> Its not dividable by {2,3,5,7,11,13} => PRIME=> ACCEPTED.
213=>Divisible by 3 => Rejected
217=>Divisible by 7 => Rejected
219=>Divisible by 3 => Rejected

Hence only Prime number in the list => 211

SMASH THAT B.

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Re: How many prime numbers exist between 200 and 220?  [#permalink]

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kinjiGC wrote:
How many prime numbers exist between 200 and 220?

(A) None
(B) One
(C) Two
(D) Three
(E) Four

First, we can omit all the even numbers (since they are divisible by 2) and all the odd numbers ending in 5 (since they are divisible by 5). So we are left with 201, 203, 207, 209, 211, 213, 217 and 219. We can omit 201, 207, 213 and 219 also since all of these numbers are divisible by 3 (notice that the sum of their digits is divisible by 3). So we only need to consider 203, 209, 211 and 217.

203/7 = 29 → So 203 is not a prime.

209/7 = 29 R 6, 209/11 = 19 → So 209 is not a prime.

211/7 = 30 R 1, 211/11 = 19 R 2, 211/13 = 16 R 3, 211/17 = 12 R 7 → So 211 is a prime.

217/7 = 31.→ So 217 is not a prime.

Therefore, there is only 1 prime number between 200 and 220.

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