Question

Let f be a product of any 15 consecutive natural numbers. is f divisible by 143?

Answer 1

let the 15 consecutive nos are x, (x+1), (x+2),......x+(15-1)1 that is x, (x+1),....., (x+14)
so their product x(x+1).... (x+14) or
(x+14)(x+13)(x+12)..... (x+1)x= (x+14) !
since 143 is multiple of 11 and 13
so putting x= 0,1,2,3,4,5....
we get the values of (x+14)! as 14!,15!,16!,.....respectively
so in 14! ,15!, 16!, ....we get the factors 13 and 11
so all of these are divisible by 143
means (x+14)! is divisible by 143 for x=0,1,2,3,....