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Prime numbers (Knewton Prep)

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by RadiumBall » Fri Apr 01, 2011 9:18 pm
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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by avdhutgmat » Fri Apr 01, 2011 9:30 pm
ans is 0
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by manpsingh87 » Fri Apr 01, 2011 11:42 pm
RadiumBall 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
*p=p!; therefore *7=7!=5040; *7+3=5043 which is divisible by 3 hence not prime,
5044, divisible by 2 hence not prime,
5045, divisible by 5 hence not prime,
5046 divisible by 2 hence not prime,
*7+7=5047 divisible by 7 hence not prime.

as there is no prime no. between *7+3 and *7+7 hence A..
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by rohu27 » Sat Apr 02, 2011 1:38 am
thrs a similar gmat prep or og prb i remember.

Q asks number of primes betwn 7!+3 and 7!+7
7!+3=3[(1*2*4*5*6*7)+1]
similarly all others can be re written as multiples of 4,5,6 and 7
as we can see each is a multiple of some number it cannot be a prime.
as its 7! we can even calculate 7! but the above method is useful for higher factorials.
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by ccassel » Sat Apr 02, 2011 5:06 pm
rohu27 wrote:thrs a similar gmat prep or og prb i remember.

Q asks number of primes betwn 7!+3 and 7!+7
7!+3=3[(1*2*4*5*6*7)+1]
similarly all others can be re written as multiples of 4,5,6 and 7
as we can see each is a multiple of some number it cannot be a prime.
as its 7! we can even calculate 7! but the above method is useful for higher factorials.

Under this method, would it be a correct assessment to say if there was a remainder (+ a value) at the end of this equation that could not be factored out we would need to multiply the n! and add the remainder to determine if it is a prime number or not? Or is there another shortcut for this scenario?

For example, 7! + 11

btw - great shortcut!
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by RadiumBall » Sat Apr 02, 2011 8:42 pm
I posted this problem just to know if there any such shortcut.

Thanks anyway, OA: A
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by Strongt » Sat May 28, 2011 3:35 am
I can find the answer by doing all the calculations, but this will take forever. Is there a shortcut for this type of questions?
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by cans » Sat May 28, 2011 4:20 am
rohu27 wrote: Q asks number of primes betwn 7!+3 and 7!+7
7!+3=3[(1*2*4*5*6*7)+1]
similarly all others can be re written as multiples of 4,5,6 and 7
as we can see each is a multiple of some number it cannot be a prime.
as its 7! we can even calculate 7! but the above method is useful for higher factorials.
This is the shortcut...
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by sourabh33 » Sat May 28, 2011 4:41 am
Prime no -> any number that does not have a divisor other than 1 & itself

Now,

*7+3 = 3 x (1.2.4.5.6.7 + 1) -> This has more divisors other than 1 & *7+3
*7+4 = 4 x (1.2.3.5.6.7 + 1) -> This has more divisors other than 1 & *7+3
*7+5 = 5 x (1.2.3.4.6.7 + 1) -> This has more divisors other than 1 & *7+3
*7+6 = 6 x (1.2.3.4.5.7 + 1) -> This has more divisors other than 1 & *7+3
*7+7 = 7 x (1.2.3.4.5.6 + 1) -> This has more divisors other than 1 & *7+3

There no of primes = 0
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by SoCan » Sat May 28, 2011 7:10 am
Strongt wrote:I can find the answer by doing all the calculations, but this will take forever. Is there a shortcut for this type of questions?
If you add two multiples of any given number X, that sum will be divisible by X.

By definition, 7! is a multiple of 3, 4, 5, 6 and 7. Therefore, the sum of 7! and any of those five numbers is divisible by that number.

To extend the concept, adding higher multiples of any of those numbers will still result in a non-prime number.
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