xyz zyxxzyyx因为x,z都不为0,如果z>1的话xyz* zyx的首位数将不小于2x,矛盾,因此z=1则xyz=100x+y+1zyx=100+10y+xxyz*zyx=10001x+1001xy+100x^2+110y+10y^2+100x1yyx=10001x+1000+110y两式相等,有:1001xy+100x^2+10y^... 分析总结。 xyzzyxxzyyx因为xz都不为0如果z1的话xyzzyx的首位数...
xyz*zyx=10001x+1001xy+100x^2+110y+10y^2+100 x1yyx=10001x+1000+110y 两式相等，有：1001xy+100x^2+10y^2=900 两边除以10, 得：1001xy/10+10x^2+y^2=90 xy需整除10，且xy/10须为0，否则上式左边大于1001，右边为90，矛盾。故只能取y=0, x=3,等式为：301*103=31001 ...
xyz和zyx是两个三位数,xzyyx是一个五位数 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 xyz zyxxzyyx因为x,z都不为0,如果z>1的话xyz* zyx的首位数将不小于2x,矛盾,因此z=1则xyz=100x+y+1zyx=100+10y+xxyz*zyx=10001x+1001xy+100x^2+110y+10y^2+100x1yyx=10001x+1000...
xyz和zyx是两个三位数,xzyyx是一个五位数 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 xyz zyxxzyyx因为x,z都不为0,如果z>1的话xyz* zyx的首位数将不小于2x,矛盾,因此z=1则xyz=100x+y+1zyx=100+10y+xxyz*zyx=10001x+1001xy+100x^2+110y+10y^2+100x1yyx=10001x+1000...
则:xyz=100x+y+1,zyx=100+10y+x. xyz×zyx=10001x+1001xy+100x²+110y+10y²+100 xzyyx=x1yyx=10001x+1000+110y 两式相等,有:1001xy+100x²+10y²=900 两边除以10, 得:1001xy÷10+10x²+y²=90 xy需整除10,且xy÷10须为0,否则上式左边大于1001,右边为90,矛盾. ...
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integrity sha512-qgIg8BcZgd0G/Cz916D5+9kqX0c7nPZyXaP8R2tLNN5tkyIZdG5fEwBrxwplzSnjC1jvQmyMNVwUCZPcbGY7Pg== dependencies: "@babel/helper-plugin-utils" "^7.16.7" "@babel/plugin-transform-react-jsx-development@^7.16.7": version "7.16.7" resolved "https://registry.npmmirror.co...