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For any integer p, *p is equal to the product of all the int  [#permalink]

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For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *7 + 3 and *7 + 7, inclusive?

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

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Originally posted by gmatpapa on 19 Apr 2011, 04:33.
Last edited by Bunuel on 30 Sep 2013, 04:23, edited 1 time in total.
Renamed the topic and edited the question.
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Baten80 wrote:
7!+3 to 7!+4
so 5043 to 5047 [5043, 5044, 5045, 5046 and 5047]
there is no prime numbers.
Ans. A
What others easy way?

The quick way is to realize that if a factor(or any of its multiples) is added to a multiple of that factor, the result will be divisible by that factor. For example: 3 is a factor of 9. If 3 is added to 9, the result will be divisible by 3. Is 5 is added to 25, the result will be divisible by 5.

Coming to the problem. You see, 7! will be divisible by all numbers from 1 through 7. In other words, all integers from 1 to 7 are factors of 7! So, if any number between 1 to 7 is added to 7!, the result will be divisible by the number that is added (if 3 is added to 7!, result will be divisible by 3. If four is added, the result will be divisible by 4 and so on..) Essentially, the numbers between 7!+3 and 7!+7, inclusive will be: 7!+3, 7!+4, 7!+5, 7!+6, 7!+7. All these numbers will be divisible by one or the other number between 3 to 7, hence making all of them non-prime.

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7!+3 to 7!+4
so 5043 to 5047 [5043, 5044, 5045, 5046 and 5047]
there is no prime numbers.
Ans. A
What others easy way?
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gmatpapa wrote:
For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *7 + 3 and *7 + 7, inclusive?

(A) None

(B) One

(C) Two

(D) Three

(E) Four

Q: How many prime numbers between 7!+3 and 7!+7

7!+3=7*6*5*4*3*2*1+3=3(7*6*5*4*2+1). Not prime.
7!+4=7*6*5*4*3*2*1+4=4(7*6*5*3*2+1). Not prime.
7!+5=7*6*5*4*3*2*1+5=5(7*6*4*3*2+1). Not prime.
7!+6=7*6*5*4*3*2*1+6=6(7*5*4*3*2+1). Not prime.
7!+7=7*6*5*4*3*2*1+7=7(6*5*4*3*2+1). Not prime.

We can infer:
No primes between "n!+2 AND n!+n".
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Here's a similar problem:

ds-is-x-x-1-a-prime-number-61428.html
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7! = 720 * 7 = 5040

# in question = 5043, 5044, 5045, 5046, 5047

none of these are prime

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gmatpapa wrote:
For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *7 + 3 and *7 + 7, inclusive?

(A) None

(B) One

(C) Two

(D) Three

(E) Four

Here's my thinking (pundits please correct me if I'm wrong):
Generally *p or p! will be divisible by ALL numbers from 1 to p. Therefore, *7 would be divisible by all numbers from 1 to 7.

=> *7+3 would give me a number which is a multiple of 3 and therefore divisible (since *7 is divisible by 3)
In fact adding any "prime" number between 1 to 7 to *7 will definitely be divisible.

So the answer is none (A)!

Supposing if the question had asked for prime numbers between *7 + 3 and *7 + 11 then the answer would be 1. For *7 +3 and *7 + 13, it is 2 and so on...
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1
7! = 7*6*5*4*3*2*1 = 42* 120 = 5040
Numbers in question = 5043, 5044, 5045, 5046, 5047

None of these are prime as they are divisible by the numbers present in 7!

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7! + 1<a< 7 are all non prime numbers.
Hence A.
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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*p is nothing but p!

for any a>=5 *a will end with 0
so +3 will give xxx3
and +7 will give xxx7
between 3 and 7 there is only 5 which cant be a prime number.
so ans None
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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gmatpapa wrote:
Baten80 wrote:
7!+3 to 7!+4
so 5043 to 5047 [5043, 5044, 5045, 5046 and 5047]
there is no prime numbers.
Ans. A
What others easy way?

The quick way is to realize that if a factor(or any of its multiples) is added to a multiple of that factor, the result will be divisible by that factor. For example: 3 is a factor of 9. If 3 is added to 9, the result will be divisible by 3. Is 5 is added to 25, the result will be divisible by 5.

Coming to the problem. You see, 7! will be divisible by all numbers from 1 through 7. In other words, all integers from 1 to 7 are factors of 7! So, if any number between 1 to 7 is added to 7!, the result will be divisible by the number that is added (if 3 is added to 7!, result will be divisible by 3. If four is added, the result will be divisible by 4 and so on..) Essentially, the numbers between 7!+3 and 7!+7, inclusive will be: 7!+3, 7!+4, 7!+5, 7!+6, 7!+7. All these numbers will be divisible by one or the other number between 3 to 7, hence making all of them non-prime.

Agree to this.... there is no need to "actually calculate" the factorial & sum up.

The factorial part (7!) has factors of all numbers stated from 3 to 7, inclusive. So they are indeed not prime
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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hi
is *p equal to p!??? becoz the question says the product of the integers between 1 and p and hence it does not include p. just a doubt...

siva
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For any integer p, *p is equal to the product of all the int  [#permalink]

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sivaspurthy wrote:
hi
is *p equal to p!??? becoz the question says the product of the integers between 1 and p and hence it does not include p. just a doubt...

siva

No, the question mentions that "*p is equal to the product of all the integers between 1 and p, inclusive" and hence *p =1*2*3...p = p!

Hope this helps.

P.S.: nice display pic! 2003 world cup I think.
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GRE 1: Q169 V154 Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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Using the rule => multiple + multiple = multiple
we can say that 7!+4,7!+5,7!+6 are all non primes.
hence A
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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gmatpapa wrote:
Baten80 wrote:
7!+3 to 7!+4
so 5043 to 5047 [5043, 5044, 5045, 5046 and 5047]
there is no prime numbers.
Ans. A
What others easy way?

The quick way is to realize that if a factor(or any of its multiples) is added to a multiple of that factor, the result will be divisible by that factor. For example: 3 is a factor of 9. If 3 is added to 9, the result will be divisible by 3. Is 5 is added to 25, the result will be divisible by 5.

Coming to the problem. You see, 7! will be divisible by all numbers from 1 through 7. In other words, all integers from 1 to 7 are factors of 7! So, if any number between 1 to 7 is added to 7!, the result will be divisible by the number that is added (if 3 is added to 7!, result will be divisible by 3. If four is added, the result will be divisible by 4 and so on..) Essentially, the numbers between 7!+3 and 7!+7, inclusive will be: 7!+3, 7!+4, 7!+5, 7!+6, 7!+7. All these numbers will be divisible by one or the other number between 3 to 7, hence making all of them non-prime.

Great Explanation!

let me add it up of my own explanation...

so the list is
(7!+3), (7!+ 4), (7! + 5), (7! + 6), (7! + 7)

Notice that 7! = 7 x 6 x 5 x 4 x 3 x 2 (there is 5 and 2 in there so the Unit Digit must be 0)

is one of (7!+ 4), (7! + 5), (7! + 6) a prime number ?

just focus on the unit digit ! a prime number is a number that is divisible only by 1 and its own number,

unit digit of (7!+ 4) is 4 (since unit digit of 7! is 0), so therefore it is divisible by 2 (because it is even) = NOT A PRIME NUMBER
unit digit of (7! + 5) is 5, so therefore it is divisible by 5 = NOT A PRIME NUMBER
unit digit of (7! + 6) is 6, so therefore it is divisible by 2 (because it is even) = NOT A PRIME NUMBER

There you have it !
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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gmatpapa wrote:
For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *7 + 3 and *7 + 7, inclusive?

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

Here is how I did it.

If we interpret *p to be multiple of all prime numbers instead of all numbers, it might be a touch easy.

*7 + 3 = 1 * 2 * 3 * 5 * 7 + 3 = 213
*7 + 7 = 1 * 2 * 3 * 5 * 7 + 7 = 217

214, 215 and 216 as we know are not prime.
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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Hello Bunuel,

The "Similar problems" links are of great help. Is there any place where I can find a problem of a certain type and then similar problems so that I can practice all variations of that concept/problem type?
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Re: For any integer p, *p is equal to the product of all the int  [#permalink]

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makshinde wrote:
Hello Bunuel,

The "Similar problems" links are of great help. Is there any place where I can find a problem of a certain type and then similar problems so that I can practice all variations of that concept/problem type?

You can check categorised questions in our questions bank: https://gmatclub.com/forum/viewforumtags.php
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_________________ Re: For any integer p, *p is equal to the product of all the int   [#permalink] 03 May 2020, 07:27

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