If the required number is divided by 3 and the remainder is 2, then 70× 2 = 140, 140 is a number that can be divisible by 5 and 7 as well as by 3, and the remainder is 2.
If the required number is divided by 5 and the remainder is 3, then the number 2 1× 3 = 63, 63 is a number that can be divisible by 3 and 7, and can be divided by 5, and the remainder is 3.
If the required number is divided by 7 and the remainder is 2, then the number 15×2=30, and 30 is a number that can be divisible by 3 and 5, and then divided by 7, and the remainder is 2.
In addition, 140+63+30 = 233, because both 63 and 30 can be divisible by 3, so the remainder of 233 and 140 divided by 3 is the same, and they are both residues 2. Similarly, the remainder of 233 and 63 divided by 5 is the same, both of which are 2. So 233 is a number that meets the requirements of the topic.
The least common multiple of 3, 5, and 7 is 105, so the remainder divided by 3, 5, and 7 will not change after the integer multiple of 233 plus or minus 105, so the obtained number can meet the requirements of the topic. Because the demand is only the number of a small group of soldiers, that is to say, the number of soldiers does not exceed 100, so 233 MINUS twice of 105 to get 23 is the demand.
Or;
It becomes a pure mathematical problem: there is a number, the remainder 2 is divided by 3, the remainder 3 is divided by 5 and the remainder 2 is divided by 7. Find this number.
It is known that 2 1m+2 = 5n+3 = x, that is, 2 1M=5N+ 1M, where n is an integer.
From the above formula, we know that any product with a mantissa of 2 1 or 6 is the solution of the above formula, so there are m = 1, 6,1,16, 2 1, 26, 3/kloc. ...
Then the corresponding x is x = 2 1 m+2 = 21(5k+1)+2, k = 0,1,2, 3, 4. ......
Then 23 128233 ... that is, the arithmetic series 23+ 105K(K=0, 1, 2, 3 ...) is its solution.