4. Find 2019 consecutive composite integers.

Question

Question

Answer 1

Answer #1

(2020)! + 2

(2020)! + 3

(2020)! + 4

(2020)! + 5

(2020)! + 6

..................

(2020)! + 2020

Note that here we have 2019 consecutive integers.

The, nth integer is of the form, (2020)! + (n+1), where, 1 $\leq$ n $\leq$ 2019

So, we can take the factor (n+1) common from the term (2020)!+(n+1) can get composite numbers for all n

Hence, each number is composite and they are consecutive as well and we have listed 2019 numbers.

Note : 2020! = 1•2•3•4•....• 2020, so, it has all the factors starting from 2 to 2020

Answer 2

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