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

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Joined: 12 Aug 2015
Posts: 2430
GRE 1: 323 Q169 V154
Re: For any integer p, *p is equal to the product of all the int [#permalink]

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14 Mar 2016, 04:48
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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23 Apr 2016, 00:57
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.

Answer A.

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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21 Feb 2017, 05:17
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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31 Mar 2017, 23:37
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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Posts: 43898
Re: For any integer p, *p is equal to the product of all the int [#permalink]

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01 Apr 2017, 04:57
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] 01 Apr 2017, 04:57

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

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