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Which of the following numbers is not prime ? [#permalink]

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03 May 2012, 01:14

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Which of the following numbers is not prime? (Hint: avoid actually computing these numbers.)

A. 6!-1 B. 6!+21 C. 6!+41 D. 7!-1 E. 7!+11

Could please someone expalin the logic behind this ? Even though I picked the right answer while solving I am not pretty clear about the underlying concept?

Which of the following numbers is not prime? (Hint: avoid actually computing these numbers.)

A. 6!-1 B. 6!+21 C. 6!+41 D. 7!-1 E. 7!+11

Could please someone expalin the logic behind this ? Even though I picked the right answer while solving I am not pretty clear about the underlying concept?

Thanks, Abhi

Notice that we can factor out 3 out of 6!+21 --> 6!+21=3*(2*4*5*6+7), which means that this number is not a prime.

Re: Which of the following numbers is not prime ? [#permalink]

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03 May 2012, 01:36

Bunuel wrote:

abhi47 wrote:

Which of the following numbers is not prime? (Hint: avoid actually computing these numbers.)

A. 6!-1 B. 6!+21 C. 6!+41 D. 7!-1 E. 7!+11

Could please someone expalin the logic behind this ? Even though I picked the right answer while solving I am not pretty clear about the underlying concept?

Thanks, Abhi

Notice that we can factor out 3 out of 6!+21 --> 6!+21=3*(2*4*5*6+7), which means that this number is not a prime.

Answer: B.

@ Bunuel - what inference does 'factor out 3' make? can we say that the second part of the options (11,41) are prime so resultant could be a prime? but 1. could u pls explain?

Which of the following numbers is not prime? (Hint: avoid actually computing these numbers.)

A. 6!-1 B. 6!+21 C. 6!+41 D. 7!-1 E. 7!+11

Could please someone expalin the logic behind this ? Even though I picked the right answer while solving I am not pretty clear about the underlying concept?

Thanks, Abhi

A prime number has only two factors - 1 and itself.

Without calculating, we cannot say whether 6!-1 or 6!+41 will be prime.

But, I can say that 6!+21 will not be prime. The reason is that 6!+21 = 3(1*2*4*5*6 + 7) (taking 3 common). This means that whatever, the value of 6!+21, it can be written as the product of two numbers: 3 and something else. Hence, this number, 6!+21, definitely has 3 as a factor and hence it cannot be prime. Since a PS question can have only one correct answer, we don't have to worry about the other options. We can say with certainty that they must be prime.
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Re: Which of the following numbers is not prime ? [#permalink]

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12 Jun 2013, 15:26

Prime numbers are of the form 6n+1 or 6n-1. The first part of each of the terms contains a 6,and hence is a multiple of 6. We only need to factor out 6's from the 2nd part of each option. If you're left with a number greater than one, then that's the answer . In this case, that would be B.

Prime numbers are of the form 6n+1 or 6n-1. The first part of each of the terms contains a 6,and hence is a multiple of 6. We only need to factor out 6's from the 2nd part of each option. If you're left with a number greater than one, then that's the answer . In this case, that would be B.

Don't get your solution...

The property you are referring to is: any prime number \(p\) greater than 3 could be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.

That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.
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Re: Which of the following numbers is not prime ? [#permalink]

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12 Jun 2013, 18:52

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This post was BOOKMARKED

Bunuel wrote:

Don't get your solution...

The property you are referring to is: any prime number \(p\) greater than 3 could be expressed as \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.

That's because any prime number \(p\) greater than 3 when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

But: Note that, not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number.

Thank you! Glad I can correct my understanding now rather than later.

A prime number can have only two factors:1 and itself. If we are able to prove that a number is divisible by any other number too, then we can say that the number is contention is not prime.

In this particular question, checking by options we can certainly say that 6! +21 = 3(6*5*4*2*1 + 7) Hence 6! +21 is divisible by 3 and thus not a prime number
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Which of the following numbers is not prime ? [#permalink]

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22 Sep 2015, 09:18

Not sure if my logic is correct, but the way I figured this out was:

Rule: We know that all primes above 3 are in the form of either 6n-1 or 6n+1.

So:

A. 6!-1 -- Here we have 6*5*4,etc -1 (thus, in the form of 6n-1) B. 6!+21 -- Here we have 6*5*4,etc + 7*3 -> This could be the answer as 21 isn't a prime number C. 6!+41 -- Here we have 6*5*4,etc + prime number D. 7!-1 -- Here we have 7*6*5*4,etc - 1 (thus, in the form of 6n-1) E. 7!+11 -- Here we have 7*6*5*4,etc + prime number

Not sure if my logic is correct, but the way I figured this out was:

Rule: We know that all primes above 3 are in the form of either 6n-1 or 6n+1.

So:

A. 6!-1 -- Here we have 6*5*4,etc -1 (thus, in the form of 6n-1) B. 6!+21 -- Here we have 6*5*4,etc + 7*3 -> This could be the answer as 21 isn't a prime number C. 6!+41 -- Here we have 6*5*4,etc + prime number D. 7!-1 -- Here we have 7*6*5*4,etc - 1 (thus, in the form of 6n-1) E. 7!+11 -- Here we have 7*6*5*4,etc + prime number

So answer B

Hi IgnacioDeLoyola,

The relation that you wrote is correct, but its reversal is not true. (you are assuming this in options A and D) All prime numbers are of the form 6n+/- 1 but all numbers of the form 6n+/- 1 are not prime Example: 7 = 6(1)+1 - Prime 25 = 6(4)+1 - Not Prime

Also, it is not necessary that any number when added to a prime number will be prime (which I feel you are assuming in options C and E) Example: 7 (prime) +3 (prime) = 10 (not prime)

You should rely on finding the factors if you have to identify that a number is prime or not.
_________________

Re: Which of the following numbers is not prime ? [#permalink]

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24 Dec 2016, 10:39

mvictor wrote:

I got it really fast 6! is not a prime, so in order to get a non-prime number, we have to add a non-prime number. 21 is not a prime number, therefore 6!+21 is not prime.

Don't get ur solution: Are we trying to say that adding two prime numbers will give a prime number?

Re: Which of the following numbers is not prime ? [#permalink]

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03 Apr 2017, 23:55

A prime no. will have only two factors 1 and itself.

Adding prime nos. can give a prime (2+3=5) or a composite no. (3+5=8) (composite no. means any number which is not prime as it has prime factors>2 i.e. it is composed of other prime factors such as 12= 2^2*3).

Back to the question:

Only 6! +21 can be written as a product of more than 2 nos. including 1.

1*2*3*4*5*6 + (3*7) =3(1*2*4*5*6+ 7)

Thus its not prime.
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Help me make my explanation better by providing a logical feedback.