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1a:Tf28,ある程度の数学の基礎は身についていると思うのでその先の勉強方法について話したいと思います。
数学の難しい問題というのは解き方の展望が見えてこないものが多くあります。なので、正確に文章を読んで、文章の中からヒントを拾ったり、式の形をみて、使えそうな公式や、定石となる解き方を考えてみることが必要になります。おそらくランボさんはこのようにして、いくつか選択肢に上がった解法の中に正解となる解法があったのにそれが使えなかった、ということだと思います。しかし解き方を思いついてから最終的な解答方針まで見えてくることはほとんどないと思います。難しい問題はイメージとしては壁が2〜3段階あるという感じです。最初の足がかりとなる解き方をして出てきた式が解けない。そして再び考える。それに対して解き方を考えまたやる。問題を解く時はこれの繰り返しになってきます。
難しめの問題のイメージを話したので、次は勉強方法について書いていきたいと思います。数学は多くの問題集に手を出すより、一冊完璧に、とよく言いますが、その通りだと思います。なぜなら、結局一冊の中に大方必要になってくる解法は全て入っているからです。そして例えばプラチカであればその単元ごとにまとめて学習していくことをお勧めします。その時に確率であれば、P型、C型、漸化式型、円や数珠順列、条件付き確率、じゃんけんや、勝敗を決めるパターン、etcがあると思うので、そのパターンを「漏れなく、だぶりなく」身に付けるとともに、どのパターンの問題はどうゆうような問題文になっているのかを自分なりに考察することが大切です。例えば、簡単な例ですが、組み合わせの時に同じようなものを区別するかしないかで解き方が変わると思います。このように問題文や式を観察して、どのときにどのパターンを使うことが多いか分類すると良いでしょう。このとき、「漏れ」がないことで、どれかのパターンに帰着し、「だぶり」がないことで、実は同じ解法なのに出題形式が違うから両方覚えてしまって、どっち使うか迷うような手間が省けます。そこを意識して勉強するのがいいと思います。
最後に過去問についてですが、過去問はあくまで出題形式、傾向や、時間などを確認して実践するものだと思っています。なので直近6年のものは残しておくべきでしょう。またマスターって言葉の定義は曖昧です。マスターが過去問の解き方を覚えるだけであるなら無駄だと思います。問題を見て、なんでこの解法をしたのか考え、そして始めてその問題を見たと仮定したとき、その問題文からどんなキーワードを拾ったら、自分がその解法にたどり着くかというところまで考え、身に付けることができて、始めてマスターしたと言えます。それなら過去問のマスターはかなり有用だと思います。数学は初見で考え、解いて、解答をみて、終わる人が多く、初見で考えることが重要だと思われがちですが、それを可能にするには解答をみた後の上記の考察がもっとも重要になると思います。
試験本番までまだあと4ヶ月あります。十分に身に付けるだけの時間はあると思うので最後まで頑張ってください。応援しています。1b:T1307,　確かに解法暗記は大切です。しかし、それを単純暗記で終わらせてしまっては危険です。京大の整数問題を例に見ていきましょう。

「n^3ー7n＋9が素数となるような整数nを全て求めよ。」（2018）
　この問題は、整数kを用いて、nを3k、3k＋1、3kー1とに場合分けして考えればすぐ解けます。しかし、この解法を単純に暗記しても、どこからこの解法を導く着想を得たのかが分からなければ、同じ解法を使う問題に対峙してもそれを見抜くことは困難です。この問題では、n＝1を仮に入れてみると、値は3で素数です。次に、n＝2を入れてみた場合、こちらも値は3で素数です。n＝3の場合は15で素数ではない、n＝4の場合は45で素数ではない、n＝5の場合は99で素数ではない……。ここで何か気づくでしょう。すなわち、実験して得られた値は全て3の倍数になっていることに気づくはずです。となれば、与式の取りうる値は全部3の倍数なんじゃないか？という疑いが生じるでしょう。この仮説を確かめるために、まずはすべてのnに対し与式の値は必ず3の倍数になるということを証明すればよいことになり、そのためにnを3で割った余りに注目して場合分けをするという解法に辿り着くわけです（したがって、modを使えばもっと楽な計算で証明できます）。(i)n＝3kの場合は言うまでもないとして、(ii)n＝3k＋1の場合、与式は27k^3＋27k^2ー12k＋3で、(iii)n＝3kー1の場合、27k^3ー27k^2ー12k＋15で、いずれも3の倍数になります。素数の中で3の倍数は3だけなので、結局この問題は、（与式）＝3という方程式を整数nについて解けば良いということになります。
　
　こんな感じで解法を深く見つめていくと、解ける問題も増えていきます。例えば、この問題。
「pが素数ならばp^4＋14は素数でないことを示せ。」（2021文系）
　p＝2のとき値は30、p＝3のとき値は95、p＝5のとき値は639、p＝7のとき値は2415、p＝11のとき値は14655……。p＝3のとき以外は、いずれも3の倍数です。よって、(i)p＝3のときと、(ii)p ≠ 3の時で場合分けをして、(ii)p ≠ 3のときでは、さらに(a)p ≡ 1（mod3）のときと、(b)p ≡ 2（mod3）のときとで場合分けして、p^4＋14が素数pに対し常に3の倍数となることを証明し、そのとき取りうる値は3のみであるが、p^4＋14はp＝2で最小値30であるから、3を取ることはない。したがって、p^4＋14は素数ではない、という解決ができるわけです。
　
　また、この問題も。
「素数p, qを用いて、p^q＋q^pと表される素数をすべて求めよ。」（2016理系）
　pとqの対称性からp≦qとしても一般性は失われないので、この大小関係のもと進めていきます。まず、2数の偶奇が一致するとき、その和は必ず偶数になりますが、pとqはいずれも素数なので、与式の取りうる値は最小でも8（p＝2, q＝2）であり、値が2となることはありません。このことから、与式の値は奇数であり、そのためにはp＝2でなければなりません（片方は偶数でなければならず、p^qが偶数となるのはp＝2の場合だけ）。すると、p＝2と固定して、qに3、5、7、11……と入れてみればいいわけです。q＝3のとき値は17で素数、q＝5のとき値は57で素数ではない、q＝7のとき値は177で素数ではない、q＝11のとき値は2169で素数ではない……。q＝3のときを除いて、すべて3の倍数ですね。しかし、この問題では、安易にqを3で割った余りで場合分けしてもうまくいきません。場合分けにさらなる工夫が必要になりますが、そこは自力でやってみましょう。

　上の問題は、いずれも同じところから解法の着想を得ていることがわかったと思います。と同時に、個別の問題にだけ通用するような覚え方をしても、似た問題ですら手が止まってしまうということも。やはり何事も、勉強というからには自分の頭で考えなければなりません。ただ単に、与えられた結果の知識や表現を覚えるだけではダメですね。その点、受験勉強は大変なものですが、そういったことも志望校という目標に向かって一途に続けられる人こそ、本番で勝っていく人たちなのでしょう。私も偉そうなことは言えませんがね。1c:T13f3,こんにちは！RIZと申します。
問題集の問題は解けるけれど初見の問題では解けなくなるということですね。
まずとても当たり前の話をしますが、数学は問題文から解答を考えなければなりません。現在の、問題集の問題は解けるけれども初見の問題では手が止まってしまうというのは、単に問題集の答えを覚えているだけに他なりません。そこで、今回は初見の問題でも解けるようにするためにはどのようにすれば良いかについてお話しします。
前提として、数学の公式や定義はしっかり学習しているとします。もし質問文に書かれている数学用語というのがこうした公式や定義であるなら、定義はまずしっかり覚えてください。そして公式についてはできれば丸暗記するより、導出できるようにしたほうが良いです。ただもう時間があまりないので最悪丸暗記でもいいですが、導出できるようにすることで、なぜその公式が成り立つのか理解できるので覚えやすくもなりますし、もし忘れてしまっても対応できるようになるのでおすすめです。例えば三角関数の2倍角とか3倍角なんかは加法定理とか、数3ですがド・モアブルの定理などから簡単に導出できますよね。加法定理を毎回導出するのは流石に面倒ですが、2倍角や3倍角を加法定理から導出するのは少しの時間でできますよね。このようにあまり覚えていなくても簡単に導出できる公式はなるべく導出できるようにした方が良いです。
さて、話を戻しますが、以上のように公式や定義が頭に入っていることを前提として、初見の問題でどのように対処するべきかについてお話しします。まず冒頭でもお話ししたように、数学は問題文だけから解答を考えなければなりません。そこでまず、問題文の条件に着目します。条件というのはいろいろあります。例えばnを自然数とするとか、x、yが円の方程式を満たしているとか、垂直に交わるとか、さまざまです。他にも、直接的には書かれていないけれども重要な条件もあります。例えば与えられた式が対称式であるとかです。こうした条件から、解答を考えていきます。例えば上の例で言えば、nを自然数として、かつnに関する命題が与えられて証明しなさいといった問題であれば、自然数かつ証明問題であることから数学的帰納法が浮かびますし、x、yが円の方程式を満たしていて、かつx、yの2変数からなる関数の最大最小を考えたい時、xとyが円の方程式を満たすという条件から、θを媒介変数としてx、yをcosθとsinθで置くとかが考えられます。他にも、垂直に交わるという条件があれば、例えばその垂直に交わる直線の傾き同士の積は−1とか、内積0とか、あるいは図形的に三平方の定理を利用することも可能かもしれません。以上のように、条件を見たときにいろいろなことが考えられるようになることで、初見の問題で同じような条件が出てきたときに対応できます。もちろん入試問題というのは問題集には載っていない初見の問題である場合がほとんどです。なので普段解いている問題と全く同じでないのは当たり前ですが、条件に関して言えば部分的に共通していますよね。なのでこうしたことが想起できるようになれば、初見の問題でも対応できるようになるわけです。しかしこのように、条件を見てそこから解法を想起するというのは初見では無理ですよね。それを問題集から学ぶわけです。つまり、ただ問題を解いて、解けなかったら答えを見て覚えて終わりではなく、解法を見たとき、それが「なぜ」そうなるのかを考えます。そして、もし自分が初見でその問題を解くとしたら、まず問題文のどの条件に着目するのかを考えます。このようにすることで、解法のストックを増やしていくわけです。とにかく、解答を見たものでも初見だったらどうするのか、そして「なぜ」そうするのかまで説明できるようになることで、初見の問題でも、それまでストックした解法の引き出しから解法を想起でき、対応できるようになるわけです。なのでまずは今までやった問題集で、問題文のどの条件に着目して、「なぜ」その解答になるのか考えながら学習するようにしてみてください。以上になります。ご質問などありましたらコメント欄の方でお願いします！2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=4LlNmA6jjsEv8EDw1Kwi","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"計算の仕方"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",0,") コメント(",0,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"6/5 22:18"}]]}],["$","div",null,{"className":"coach-mark 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mb-4","children":"すべての回答者は、学生証などを使用してUniLinkによって審査された東大・京大・慶應・早稲田・一橋・東工大・旧帝大のいずれかに所属する現役難関大生です。加えて、実際の回答をUniLinkが確認して一定の水準をクリアした合格者だけが登録できる仕組みとなっています。"}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap mb-4","children":[["$","div","advice-part-0",{"children":[null,"どのような流れでその値が出たのかが分からないので、そのまま答えることしか出来ないのですが、√29 × √41 = √1189 なので、するとしたら分母の有理化をして、-22.8√1189/1189になりますね。\n\n√29ではなく、√19ならば、22.8=0.12×19なので、-0.12×√19/√41 = -0.12×√779/41になります。\n\nどれもしっくりと来ませんね…力になれずすみません。"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":null,"adviserName":"Souya","adviserDepartment":"東京大学理科一類","adviceId":"4LlNmA6jjsEv8EDw1Kwi","numberOfFan":1,"clipsAvg":2,"adviceRateAvg":5,"profile":"2024年度、現役で東大に入りました。地方の偏差値65程度の中高一貫校出身ですので、似た境遇の方は参考にできるかもです！ファンの登録とクリップよろしくお願いします〜！！"}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink パンフレットバナー画像","className":"mt-4 rounded"}]}]}],["$","div",null,{"className":"pt-4","children":["$","$L17",null,{"id":"adsbygoogle-init-under-advice"}]}]]}],["$","div",null,{"className":"flex 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mb-1","children":"はじめまして、ご質問にお答えさせていただきます、東京大学理科I類の者です。\n\n10^-1の場合はわざわざその表記にせず、0.112のように書いてあげれば良いです。\n\n有効数字は誤差を含みながらも、知りたい位まで(例えば実験などで機械が読み取れるであろう数値の限界)示れば良いです。\n\nそれを1.12×10^-1と答えようが、0.112と答えようがそこで点数が引かれるということは、大学入試においてはありえません。あるとすれば、表記の指定がある場合なので問題文はしっかり読んだほうがいいです。\n\nただ、基本的に問題文などに出てくる数値の表記に合わせてあげれば大丈夫なので、心配だというのであれば、合わせて書けばよろしいかと思います！\n"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 3.31-2 6-2m0-12C9.79 4 8 5.79 8 8s1.79 4 4 4 4-1.79 4-4-1.79-4-4-4zm0 10c-2.67 0-8 1.34-8 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"数ⅲで積分 焦ってます。"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"まずは基本的な計算が出来るようになりましょう。\nその為にも教科書の問題で良いので、部分積分や置換積分の章の問題をこなして「部分積分と置換積分のどちらをすれば良いか分かれば計算ができる」という状態にします。\nそしてここからが多くの人が悩む、どこをどう置換するのか？いつ部分積分や置換積分をするのか？という問題です。\n基本的に置換積分や部分積分の目的は複雑な関数の積分を、既知の積分に置き換えるor変形するという事です。上の計算問題をこなしていく内に、どんな形の積分なら計算できるか感覚的に分かると思います。\nそのように解ける形の積分を自分なりに頭の中で整理してどの形に変形できるかな？と考えます。\n基本的には難しい所や邪魔なところを置換するorxのべき乗やlogを消したいから部分積分する…といった理解でも正直問題はないです。中には双曲線関数など数学的に重要な例や深い視点に立てば自然な置換も存在しますが、これは誘導が着くはずなのであまり気にしなくて良いです。\n纏めると、まずはきちんと計算が出来るようになる事と計算が出来る積分形を自分なりにしっかり理解する事が肝要だと思います。\n拙い文章ですが、役に立てれば幸いです。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"この数学の問題を教えて下さい🙇"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"自然数を8で割った余りは0〜7になるのは理解できると思います。\nそこで、nを自然数とすると、\n8で割った余りが\n0→8n\n1→8n 1\n2→8n 2\n3→8n 3\n4→8n 4\n5→8n 5\n6→8n 6\n7→8n 7\nとすることですべての自然数を表すことができます。問題で聞いているのは平方数ということなので、それぞれを2乗すると、\n\n0→64n^2=8×8n^2\n1→64n^2 16n 1=8(8n^2 2n) 1\n2→64n^2 32n 4=8(8n^2 4n) 4\n3→64n^2 48n 9=8(8n^2 6n 1) 1\n4→64n^2 64n 16=8(8n^2 8n 2)\n5→64n^2 80n 25=8(8n^2 10n 3) 1\n6→64n^2 96n 36=8(8n^2 12n 4) 4\n7→64n^2 112n 49=8(8n^2 14n 6) 1\n\nとなります。\nすべて(8n ○)^2という式になる以上、n^2とnの係数は8の倍数になるので、自然数部分である余りの2乗部分を8で割った時の余りが平方数の余りになります。\n\n長くなってすみません。わからなかったらまた質問してください。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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mb-1","children":"思考を形に残すのです！\nまず京大の解答用紙はa3で、右半分が計算スペースとなっています。\nダメな人は頭の中で全て考えてごちゃごちゃと計算用紙に計算していますが、少しでも詰まると何もわからなくなり、思考停止してしまいます。\nこれは、人間の脳の特性による物です。人間は英単語などの外部に存在する物事を覚えるのは得意ですが、思考自体を記憶することができないと言う物です。\nあなたも10秒前に頭の中に考えてたことを完璧に紙に書き起こすことはできないでしょう。\n思考を残すために、最初から回答欄に解き進めると良いでしょう。\n計算のみ右の計算スペースに残すのです。\nさて、その解き進める解放についてですが、まず問題を見れば幾らかの解放が選択肢としてあると思います。\n整数ならMODや積の形にしたり連続関数のグラフで考えたり、、、、などですそのそれぞれの解法について、その問題にはどれが適しているか妥当性を吟味する癖をつけましょう。\n解けない方法で無理に考えても、何も生まれません。\n解法を絞ることができたら、解答欄で解き進めましょう。\nどんな感じかのイメージですが、YouTubeで東大理科三類のルシファーさんが数学実況をしている感じでやりましょう。\n解答欄に書きながら計算だけ違うスペースでします。\nすでにその解放に絞られているので、行き詰まった場合はその原因は計算ミス以外にありませんので計算を再度やり直すだけで解決します。\n\nこの方法は最初は1ヶ月ぐらい慣れるのにかかりましたが、慣れると5完半は余裕です。\n正直6完も行けるようになりますが、計算ミスが怖いので僕は5完半で止めて計算見直してました。2020年の過去問でも5完半ですし、今年の2024の本番でもそうでした。\n本当に安定します。\nさて、ここからの勉強の話です。\n解法をまず知ると言うことと、その解放を使いこなす必要があります。\nその二つとも同時にカバーできるのが、駿台の実践過去問集の青い本です。\nメルカリで古いものを探すなどして、2000年前後の物まで全て150分測って取りかかりましょう。\n駿台の実践模試は解放を使いこなすことができたら容易に回答できる、使いこなせないなら全く回答ができないという感じですので本当に力がつきました。\nそれが終われば京大の過去問、そして余裕があれば東大の過去問をやりましょう。\n\n後輩になってくれるのを心より楽しみにしています！"}],["$","div",null,{"className":"flex 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mb-1","children":"初めまして。rockyyyと申します。\n数学についての勉強法についてお答えします。\n\n結論から言うと、YNUさんの勉強法は間違ってはいません。何度も解き直して、解法を落とし込むという方法はとても重要です。しかし、時間がかかりすぎてしまうため、時間が惜しい受験期間においてはあまり望ましくないのかなと思いました。\n\n僕は、数学は解答を丸覚えするというよりも、なぜ、その解き方をするのか理解しながら覚え込むということをしたらいいのではないかと思います。例えば、解き方がわからなくて、解答を読んでいるときに「なぜこの計算をしているのだろう」とか、「なぜこの式変形をするのだろう」などを考えるということです。そして、その意味や理由がわかったとき、数学という教科の本質の理解に一歩近づくのではないかと思います。\n\n解法だけを覚え込んでしまうと、なぜこのような計算や操作をしたのかということがよくわからないままなので、少し問題が変わると手も足も出なくなってしまいます。\n\n具体的に、家で数学を勉強するときのおすすめの勉強法としては、第一に、なぜ解答ではこのようなことをしているのかを考えます。そして第二に、それが分かったとき(つまり、これを求めるためにこんなことをしたのか！となったとき)は、「あーはいはい。これをこうするために、こうしたわけね！」などと独り言を言うといいと思います笑。意外と記憶に残ったりします。あと、そのわかったことをノートに目立つように殴り書きなどをしておくと良いと思います。そのノートを日常的に見直していたりすると、着実に力はつくと思います。僕はそうしてました。これで数学は得意になったと思うので、間違ってはいないのではないかなと思います。(あくまで個人の意見です)\n\n数学の問題を解くときは、山登りと一緒です。山頂を目指すためのルートはたくさんあります。要は登り切ることができれば良いのですから、ルートはどれを選んでもいいわけです。つまり、そのルートを学ぶ(これは先述の、なぜこうした計算や式変形をしたのかを学ぶことと同義)ことが大切です。\n\nそれさえあれば、例え問題がかわっても、大丈夫なのではないかなと思います。要は、解答でなぜそんなことをしているのかということを理解することが重要です。\n\n今から頑張っても全然遅くはありません。よければ僕の勉強法も参考にしてもらって頑張って欲しいです！応援していますよ！"}],["$","div",null,{"className":"flex 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