What is the smallest number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7 & so on until when divided by 2 leaves a remainder of 1?

Answer 1

the smallest number I got was:

2519

Answer 2

I don't believe there's a smaller one than that (I checked it in Excel). The next one is 5039, which is twice the last plus 1. (Incidentally, the one before 2519 corresponds to N=-1.) To work out which number gives a remainder of 1 for all the integers from 1 to P, work out the least common multiple of all the numbers from 2 to P, and subtract 1. So for P=10, the LCM of 2, 3, 4, 5, 6, 7, 8, 9 and 10 is 2x3x2x5x7x2x3=2520; subtract 1 and you get your 2519.