The sum of two positive integers is 2310. Can their product be divisible by 2310?

Question

Find the largest possible value of positive integer N, such that N! can be expressed as the product of (N-4) consecutive positive integers?

Answer 1

No is the answer

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Given: a+b=2310
Let a*b divisible by 2310: a*b=2310*k.
2310 is made up of 5 different primes: 2310=2*3*5*7*11. a*b=2*3*5*7*11*k. If k ≥ 2, then a+b = a*b ≥ 2*2310. It means contradiction and it is not possible to have 2310 ≥ 2310 * 2.