From the diagram 1, we get a=3m+ 1, a∈N, m∈N, ①.

A=5n+2, a∈N, n∈N, ② are obtained from Figure 2.

From fig. 3, a=7k+3, a∈N, k∈N, ③ are obtained.

That is, the positive integer solutions of congruence groups A ≡ 1 (modulo 3), A ≡ 2 (modulo 5) and A ≡ 3 (modulo 7),

Lcm= 105。

∴a=70× 1+2 1×2+ 15×3- 105i

= 167- 105i，i∈Z，

Amin =52,

Off-topic: a= 105s+52, s∈N

Note: 70 is the smallest number divisible by 5 and 7, but divisible by 3:1;

2 1 is the smallest number that can be divisible by 3 and 7 and 5.

15 is the smallest number that can be divisible by 3 and 5 and 7.

105 is the least common multiple of 3, 5 and 7.

N is a set of natural numbers,