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English Senior High

分かる範囲で良いので解いていただけませんか、、? 答えが配られなくて、授業で必ず指されるので合ってるかどうか不安です( ; ; )

18 FIRST STAGE Chapter 文法・語法- 1 空所に入れるのに最も適当な語(句)を選びなさい。 1. "I like my job, but I wish I made more money." "Me, too. If I ( ), I could buy a new car." 3 had 2 do ℗ did 2. If I ( 1 am 3. If I ( have 仮定法 5. If she ( ) you, I would not accept that kind of offer. 2 have been 3 were 4. Would you have taken the job if you ( 1 knew 2 had known 9. I ( ) a camera with me I would have taken a picture of the lake. 2 had 3 had had 4 have had late, give her this message. 1 were coming 2 would come 6. If you were to fall from that bridge, it ( 1 is was 1 can't manage 3 couldn't manage 3 have known 3 should come 10. They got two free tickets to Canada; afford to go. cad 1 rather 2 but 4 have 3 would be 4 will be ) how terrible the conditions were? 4 would have known (立教大) (センター試験) Hon berusa 7. He would have become a great marathon runner, if it (ondow) for his knee problem. 1 was not 8. Thank you for the kind help you extended to me the other day. I ( alone. 1) happy to see him, but I didn't have time. 1 will have been 2 would be 3 will be eqi. 4 shall come ) almost impossible to rescue you. 4 would have been ( センター試験) (川崎医療福祉大 ) (京都産業大) OL 2 had not been 3 has not been 4 would not have been 2 can't have managed couldn't have managed 200 3 however (成城大) (同志社大) otherwise (南山大) (慶應大) 4 would have been ) they'd never have been able to (小樽商科大)

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Mathematics Undergraduate

やさしい理系数学例題3(2)整数分野の証明問題です。 模範解答の意味は理解できますが、16で割ったあまりで分類しようと考えるに至る過程がわかりません。

あり、その最大数はab である。 この定理について興味のある方は, 「ハイレベル理系数学」の例題3と演習問題 14 を参照されたい. 例題 3 正の整数a,b,cが a+b2=c2 をみたすとき,次の (1), (2), (3) を証明せよ . (1) a, b のいずれかは3の倍数である. (2) a,b のいずれかは4の倍数である. (3) a,b,cのいずれかは5の倍数である. 考え方 任意の整数は, 3m, 3m±1 (mは整数) などの形で表せる. 【解答】 (1) 任意の整数は3m,3m±1 (m∈Z) のいずれかの形で表せ, (3m)2 = 0, (mod3) (3m±1)²=1. よって, a, b がともに3の倍数でないとすると, ∫(a2+62)÷3の余りは,2 lc²÷3の余りは, 0,1 であるから, a2+b2=c2 となり矛盾. ゆえに,d2+b2=c2 のとき, a, 6 のいずれかは3の倍数である. (2) 任意の整数は 4m, 4m±1,4m+2 (mez) のいずれかの形で表せ , (4m)²=8.2m² = 0, (4m±1)²=8(2m²±m)+1=1,9, (mod16) (4m+2)^2=8(2m²+2m)+4=4. よって, a, b がともに4の倍数でないとすると, 背理 (a²+62)÷16の余りは, 2, 5, 8, 10, 13 lc²16の余りは, 0, 1,4,9 (5m)2 =0, (5m±1)' = 1, (mod5) (有名問題 ) (5m±2)²=4. よって, a,b,cがすべて5の倍数でないとすると, (終) なぜood 16 で分類しょうと 考える 光に平方数で割った余りを であるから, a+b2=c2 となり矛盾. ゆえに,a+b=²のとき, a,b のいずれかは4の倍数である. (3) 任意の整数は 5m,5m±1.5m±2(m∈Z) のいずれかの形で表せ, (終)

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