判断一个整数能否被7整除

2025-03-16 08:35:44
推荐回答(3个)
回答1:

判断一个整数能否被7整除,只需看去掉一节尾(这个数的末位数字)后所得到的数与此一节尾的5倍的和能否被7整除.如果这个和能被7整除,则原数就能被7整除.如126,去掉6后得12,12+6×5=42,42能被7整除,则126能被7整除.类似地,还可通过看去掉该数的一节尾后与此一节尾的n倍的差能否被7整除来判断,则n=
2

回答2:

∵和的时候,是尾数的5倍,
能被7整除,
任意一个正整数写成P=10a+b,b是P的个位数.
根据已知结论,P是7的倍数等价于a+5b是7的倍数,而a+5b=a-2b+7b,
a+5b和a-2b相差7的倍数,所以它们两个同时是7的倍数或者同时不是7的倍数.
因此n=2符合要求.
∴差的时候,应是尾数的2倍,
∴n=2.
故填2.
参考资料:魔方格
求采纳

回答3:

  判断一个整数能否被7整除,只需看去掉一节尾(这个数的末位数字)后所得到的数与此一节尾的5倍的和能否被7整除.如果这个和能被7整除,则原数就能被7整除.如126,去掉6后得12,12+6×5=42,42能被7整除,则126能被7整除.类似地,还可通过看去掉该数的一节尾后与此一节尾的n倍的差能否被7整除来判断,则n=
可以这样想12(去掉后的数)-6n=7的倍数。这样的话,填2.
  望采纳!谢谢!

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