英文數(shù)學(xué)題求救,分?jǐn)?shù)類的!
英文數(shù)學(xué)題求救,分?jǐn)?shù)類的!
A find the decimal expamsion for 1/43
B show that 9 is a factor of the sum of the digits in the repetend of 1/7.1/39,1/43
C suppose none of 2,3,5 is a factor of n.Explain why 9 is a factor of the sum ofthe digits in the repetend in the repetend of 1/n
D Find the fourth smallest value of n which neither 2 nor 5 is a factor and 9 is not a factor of the sum of the digits in the repetend of 1/n
求C 和
A find the decimal expamsion for 1/43
B show that 9 is a factor of the sum of the digits in the repetend of 1/7.1/39,1/43
C suppose none of 2,3,5 is a factor of n.Explain why 9 is a factor of the sum ofthe digits in the repetend in the repetend of 1/n
D Find the fourth smallest value of n which neither 2 nor 5 is a factor and 9 is not a factor of the sum of the digits in the repetend of 1/n
求C 和
英語(yǔ)人氣：357 ℃時(shí)間：2020-05-22 09:57:31
優(yōu)質(zhì)解答
D：根據(jù)C的結(jié)論,n必須是3的冪,也就是3,9,27,81? 是81嗎能回答c就給滿意答案了，謝謝 假設(shè)1/n化成小數(shù)后 從第k+1位開(kāi)始循環(huán)，循環(huán)節(jié)長(zhǎng)度是s，即1/n=0.x1……xk a1……as a1……as ……，則1/n=x/(10^k)+(a1……as)/(10^k * 99……9)一共s個(gè)9； 于是1/n=(x*99……9+a1……as)/(10^k * 99……9)，于是可得： (x*99……9+a1……as) * n=10^k * 99……9，等式右邊能被9整除，而n不能被3整除，因此x*99……9+a1……as能被9整除； 得a1……as能被9整除； 則9也整除a1+a2+……as
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