Hi everyone, Can someone please explain the approach to solve this problem efficiently?
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Try to factorize each of those options
thanks for response! maybe it would be clearer to understand how you’d write a number that could be a prime?
n! \pm k is composite for 2 \leq k \leq n — if that’s what you’re asking. For k > n or k = \pm 1, we generally can’t guarantee anything about n! + k being prime or not
TLDR
20! has 10, 11 and 13 in itself. hence for all options these numbers can be taken out as factors making number as composite number.
Detailed Explanation:
Try to expand numbers if it helps, For example,
20! = 1*2*3*4*5…*10*11*12*13*14*…*20
Notice how 20! contains 10, 11 and 13. which you can take out as factor for each option separately.
Remember Prime definition? 7 is prime because it cannot be broken into multiples of two whole numbers greater than 1.
But 14 is not prime because it can be broken into multiples of two whole numbers greater than 1 as 14 = 2*7
Similarly Option A can be broken into multiple of two whole numbers greater than 1,
20! + 10 = 10 * ( (1*2*…9*11*12*…*20) + 1) and both will be whole numbers > 1. Hence not Prime.
Same is true with other numbers,
Option B, You can take 11 as factor, and Option C you can take out 13 as factor.
So all of them are NOT prime.
Hope this will help. 