小明有a頭牛,每頭賣a元,將賺來的錢與小健分,分法是拿10元給小明,再拿10元給小健,如此分法,經過y次,由小明拿到10元後,剩餘的錢不足10元(有剩餘),求剩下多少錢? 幾天後再公佈.............小學 回應 訪客 (未驗證) Re: 平方的問題....很簡單 2005-06-29 22:00 將所有正整數以十為模分為十類 依據題目...其十位數字應為奇數 又(10n)^2=100n^2 (10n+1)^2=100n^2+20n+1 (10n+2)^2=100n^2+40n+4 (10n+3)^2=100n^2+60n+9 (10n+4)^2=100n^2+80n+16 (10n+5)^2=100n^2+100n+25 (10n+6)^2=100n^2+120n+36 (10n+7)^2=100n^2+140n+49 (10n+8)^2=100n^2+160n+64 (10n+9)^2=100n^2+180n+81 其中僅10n+4&10n+6兩組十位數字為奇數 故剩餘的錢為6圓 j7631103 Re: 平方的問題....很簡單 2005-06-30 14:15 BINGO.........
訪客 (未驗證) Re: 平方的問題....很簡單 2005-06-29 22:00 將所有正整數以十為模分為十類 依據題目...其十位數字應為奇數 又(10n)^2=100n^2 (10n+1)^2=100n^2+20n+1 (10n+2)^2=100n^2+40n+4 (10n+3)^2=100n^2+60n+9 (10n+4)^2=100n^2+80n+16 (10n+5)^2=100n^2+100n+25 (10n+6)^2=100n^2+120n+36 (10n+7)^2=100n^2+140n+49 (10n+8)^2=100n^2+160n+64 (10n+9)^2=100n^2+180n+81 其中僅10n+4&10n+6兩組十位數字為奇數 故剩餘的錢為6圓
回應
訪客 (未驗證)
Re: 平方的問題....很簡單
2005-06-29 22:00依據題目...其十位數字應為奇數
又(10n)^2=100n^2
(10n+1)^2=100n^2+20n+1
(10n+2)^2=100n^2+40n+4
(10n+3)^2=100n^2+60n+9
(10n+4)^2=100n^2+80n+16
(10n+5)^2=100n^2+100n+25
(10n+6)^2=100n^2+120n+36
(10n+7)^2=100n^2+140n+49
(10n+8)^2=100n^2+160n+64
(10n+9)^2=100n^2+180n+81
其中僅10n+4&10n+6兩組十位數字為奇數
故剩餘的錢為6圓
j7631103
Re: 平方的問題....很簡單
2005-06-30 14:15