C# 检查数字是否可以被 24 整除
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Check if number is divisible by 24
提问by Hoyo
I'd like to put up a if function, that will see if the variable is dividable by 24, if it's then it does the function else not, same logic however, I want to see if the output is a perfect number, e.g if we do 24/24 that will get 1, that's a perfect number. If we do 25/24 then it'll get 1.041 which is not a perfect number, the next perfect number will come when it reaches 48 that will be 48/24 which will get 2 that's a perfect number.
我想建立一个 if 函数，它将查看变量是否可被 24 整除，如果是，则它执行该函数，否则不会执行该函数，但是，相同的逻辑，我想查看输出是否为完美数，例如，如果我们做 24/24 会得到 1，这是一个完美的数字。如果我们做 25/24 那么它会得到 1.041，这不是一个完美的数字，下一个完美的数字会在它达到 48 时出现，这将是 48/24，它将得到 2，这是一个完美的数字。
采纳答案by Hanlet Esca?o
Use the Modulusoperator:
使用模数运算符：
if (number % 24 == 0)
{
...
}
The % operator computes the remainder after dividing its first operand by its second. All numeric types have predefined remainder operators.
% 运算符计算第一个操作数除以第二个操作数后的余数。所有数字类型都有预定义的余数运算符。
Pretty much it returns the remainder of a division: 25 % 4 = 1 because 25 fits in 24 once, and you have 1 left. When the number fits perfectly you will get a 0 returned, and in your example that is how you know if a number is divisible by 24, otherwise the returned value will be greater than 0.
几乎它返回除法的余数：25 % 4 = 1 因为 25 一次可以放入 24，而您还剩下 1。当数字完美匹配时，您将返回 0，在您的示例中，这就是您如何知道一个数字是否可以被 24 整除，否则返回的值将大于 0。
回答by Nikhil Agrawal
How about using Modulusoperator
如何使用模运算符
if (mynumber % 24 == 0)
{
//mynumber is a Perfect Number
}
else
{
//mynumber is not a Perfect Number
}
What it does
它能做什么
Unlike /which gives quotient, the Modulus operator (%) gets the remainder of the division done on operands. Remainder is zero for perfect number and remainder is greater than zero for non perfect number.
与/给出商不同，模运算符 ( %) 获得对操作数完成的除法的余数。完全数的余数为零，非完全数的余数大于零。
