Question

(**) find the smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, ... and a remainder of 9 when divided by 10.

Answer 1

Let N be such a number

N=10a+9:  N+1 = 10a+10

N=9b+8: N+1 = 9b+9

N=8c+7: N+1 = 8c+8

N=7d+6 and so on

N=6e+5

N=5f+4

N=4g+3

N=3h+2

N=2i+1

In other words, N+1 is a multiple of 2,3,...10

Hence find LCM

LCM of 1 to 10 is 2520

Hence N = 2520-1 = 2519

The smallest positive integer that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, ... and a remainder of 9 when divided by 10 is the number 2519