Which of the following numbers is not prime ?

@ 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?

The added or subtracted terms, all but one of them, is a prime number itself, that should be a sort of red flag.

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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.

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

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.

Primes cannot be written as a product of two numbers
Smash B
Stone Cold

Nothing just respect for you

6! + 21 = (6 x 5 x 4 x 3 x 2 x 1 ) + ( 3 x 7 )

Try to divide the number by 3 it is divisible.....

PS : Good going Austin Bro , keep it up !!

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

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.

2 Concepts are being Tested:


Rule 1: Multiple of K (+ or -) Another Multiple of K = A Multiple of K

Rule 2: Multiple of K (+ or -) NON-Multiple of K = NEVER equal a Multiple of K


6! =
Multiple of 6
Multiple of 5
Multiple of 4
Multiple of 3
Multiple of 2
and of course Multiple of 1

7! = all the above Multiples
and also Multiple of 7


Answer A-

6! - 1 =

Multiple of 6/5/4/3/2 - NON-Multiple of ANY of these Numbers = NON-Multiple of ANY of these Numbers

Answer is NOT Divisible by anything other than 1 (and ITSELF)

Answer B-

6! + 21 =

Multiple of 3 + Multiple of 3 (21 = 7 * 3) = Another Multiple of 3

Answer B is NOT a Prime Number because we know at the VERY LEAST the Result will be Evenly Divisible by 3. In other words, it will have 1 , 3 and itself as Factors at the very least, violating the Rule for a Prime Number.


you can check -C-, -D-, -E- using Similar Logic.

the Factorial and the Number Added or Subtracted from it will NOT SHARE Any Factors in Common (i.e., it will not be a Multiple of K +/- Multiple of K situation)


-B- is the Answer

My 20 seconds sol

Any factorial +- 1 is prime so A & C out
any no added to factorial greater than that factorial would be prime so C & E out left with B
more ever 6 factorial or 7! are very easy U can perform calculation dont be so static just choose concrete method to get answere

This is a factoring problem.

The only thing you can deduce from choices A and D is that they are not divisible by the factors of 6! and 7! respectively. This does not mean that they are prime, but makes it somewhat likely.
Looking at choice B, you can see that 6!+21 has a factor in common (3). This implies that whatever number 6!+21 is, it must be divisible by 3. Therefore, it cannot be prime.