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please help! need it quick!

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1. Fill each cell with a digit from 0 to 9( not necessarily distinct) so that for the seven multidigit numbers formed across and down, one is a perfect cube (343, already filled in for you) and the rest are perfect squares. A number may not begin with the digit 0.[3][4][3][] [] [] [][] [] [][] [] [] [] =... 顯示更多 1. Fill each cell with a digit from 0 to 9( not necessarily distinct) so that for the seven multidigit numbers formed across and down, one is a perfect cube (343, already filled in for you) and the rest are perfect squares. A number may not begin with the digit 0. [3][4][3] [] [] [] [] [] [] [] [] [] [] [] = boxes... please help i dun understand,, explain please 更新: sorry, the last one,, there are only 3 boxes .. don't fill the last one,, just want to tell their boxes >

最佳解答:

由於題目資料有限，我將會用試的方法求得答案。考慮第一直行和第三直行都是３字頭的四位平方數，第一步是嘗試找出這兩個數的組合，令第三橫行和第四橫行的三位數為平方數。（為方便起見，建議把３字頭的四位平方數和三位平方數（即５５平方至６３平方和１０平方至３１平方）列出，最好能使用計算機，不然做算草會相當痛苦）３０２５１００４４１３１３６１２１４８４３２４９１４４５２９３３６４１６９５７６３４８１１９６６２５３６００２２５６７６３７２１２５６７２９３８４４２８９７８４３９６９３２４８４１３６１９００４００９６１３０２５不能填入第一直行，因為有０，０不能作為一個數的起首。接著試把３１３６填入第一直行，留意以３和６起首的三位平方數末端的數字能否與其中一個四位數相符，發現沒有這四位數，因此３１３６不能填入第一直行，如此類推。得出的（第一直行，第三直行）組合有（３２４９，３６００），（３７２１，３３６４），（３７２１，３９６９），（３８４４，３６００）此時，每一個組合只空了兩個數字，再逐一嘗試，得出的答案如下：３４３７３９６２５６１６９

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