satgates wrote:

The question is mine; probably you will not find it in books.

Meanwhile, the solution is not that difficult.

Z=3n+1

Z=5k+3

Z=7p+5 the trick is to see that each time a quotinent is 2 more than a remainder. Therefore, add 2 to both parts and get what you need.

Z+2=3n+1+2=3n+3, divisible by 3

Z+2=5k+3+2=5k+5, divisible by 5

Z+2=7p+5+2=7k+7, divisible by 7

so, Z+2 is divisible by 3, 5, and 7.

To find the smallest positive Z, we have to subtract 2 from the LCM of 3, 5, 7.

LCM = 3*5*7, since they all are primes = 105

105-2=103 is the answer.

To find the second smallest number: 2LCM-2=210-2=208 and so on.

Do you like the question? It is one of my best.