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1a: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。1b:T10ea,受験数学にひらめきは全く必要ありません。

実際、数学者と数学の得意な高校生が、受験数学で勝負すると高校生が圧勝します（実話です）。一体何が、高校生を勝たせるのだと思いますか？

受験数学には、確かに、「ひらめきのようなもの」を要求する場面があります。特に整数問題などで顕著ですが。しかし、ほとんどの問題は、今まで身につけてきた解法で対応できてしまうんですね。

例えばですが、多変数関数 f(x,y)の最大値、最小値を求めよという問題が出たとします。（f(x,y)の中身は、例えば、x^2 3xy y^2などですね。ここではそれは本質ではないのでスルーします。）その時、方針が何通りかあるんですが、それを列挙できますか？
あるいは、図形問題に対して、どのようなアプローチを考えるべきか説明できますか？
（答えはどちらも回答の最後に載せますね）

もし1つも分からない場合や、何個かしか挙げられない時は、少し補充的な勉強をする必要があります。
問題ごとに、それを解くための最適な方針がありますね。それをメモ程度で十分なので、どんどんまとめていってください。すると、多種多様に見える問題も、スタートは必ず同じことをしていたり、何個かのパターンの方針しか使っていなかったりします。本当はこういうことを分かっていくのは、問題演習を通してだんだん培っていくべきものなんでしょうが、99％の人は出来ないでしょう。僕も全然出来ませんでしたし。

なんにせよ、こういう「解法の整理」をしていくと、全く手が付かない問題はほとんどなくなってきます。途中までは行けるようになるんですね。そして、「ひらめき」は大抵こういう場面で使うものですね。例えば最後の最後に有名不等式を使ったりなどでしょうか。しかし、これすらも、方針としてカテゴライズすることが可能です。いわゆる純粋なひらめきは、受験数学においてはあり得ないといって良いでしょう。大抵、「閃かない」時は、解法が浮かばない時です。かなり具体的な問題に帰着できましたね。
僕は、ノートの見開き1ページに、この問題が来たら、この方針がよく登場する！というフローチャートのようなものを作っていましたね。頭の中が整理されていく感じがして楽しいですよ。

ちなみに、基礎ができていないということは、多少あるにせよ直接的な原因ではなく、いくら固めたところで、成果が微々たるものしか出ないので、気をつけましょう。青チャート、フォーカスゴールド、どちらも持っている時点でフル装備なので、多少の復習はもちろん必要といえども、頑張る必要はありません。

さて、先ほどの問題、わからずじまいは良くないですから簡単に
多変数関数の最大最小問題：
・等式があればxかyに代入してそれを消去する（いわゆる文字消去）
・xかyのどちらかを定数とみなし、ただの１変数関数とみなして考える（いわゆる文字固定）
・有名不等式の利用（相加相乗平均の関係、コーシーシュワルツの不等式、三角不等式など）
・逆像法
・線型計画法
・グラフを書いて考える
Etc.

図形問題のアプローチ
・まずは初等幾何で解けないか考える。
・次に、位置ベクトルを導入することで、内積などを利用して解けないか考える。
・もし対称性の高い図形だったら、座標平面を設定するのも考える。

僕がこの解法整理についての対策を編み出し、始めたのは12月の半ばです。今なら相当早いタイミングから対策できますから、ぜひ過去問での得点をぐんぐん挙げて、自信をつけていってほしいと思います。

では、有意義な秋をお過ごしください！1c:T12ba,普遍的なことだけを説明しても中々伝わりづらいと思うので、具体的に問題を1問出しながら説明させてください！

まず前提として、応用の問題が解けるようになるためには以下のことが必要になります。(結論です)

・基本的な解法がすぐに出てくるようにする
・問題を見た時、前の問題との関連性から考えていく
・誘導に乗っていくのに慣れるのにはとにかく演習量が必要

1つ目は恐らく大丈夫だと思います。また、3つ目もこれから2次試験向けの演習を重ねるうちに「あの時の誘導に似てるなー」というような感覚で段々できるようになってくるものです。つまりは慣れです。自分自身もこれを強く感じています。最初は中々誘導に乗れず辛いかもしれませんが、まずは量をこなしましょう。

おそらく問題は2つ目です。
これは分かりやすく言うと、「こうやってやっていって…あ、(1)(2)ここで使う？」という考え方ではなく、「(1)や(2)の問題の考え方を上手く使えないかな〜」「今までやったことのある基本問題の考え方が何か使えないかな〜、あ、文章のこの部分前にやったあの問題文と似てるな〜」と言ったような、初めから誘導や基本問題などのヒントの方から答えを探っていくように考えていくことです(長くてごめんなさい)。

実際に問題を見て考えていきましょう！以下は2015年の九大の問題です。

以下の問いに答えよ。
(1)nが正の偶数のとき、2^n-1は3の倍数であることを示せ。
(2)pを素数とし、kを0以上の整数とする。2^(p-1)-1=p^kを満たすp,kの組を全て求めよ。
(※^の後は指数を表します。2^n-1は2のn乗-1、2^(p-1)-1は2のp-1乗-1です)

(1)は割愛しますが、n=2l(lは自然数)とかと置いて二項定理で分解して3で括ったり、帰納法を使えばいいと思います。とにかく2^n-1が3の倍数だと分かればいいです。

問題は(2)ですね。先程言った通り、誘導を上手く使えないかという点からとにかく問題を見ましょう！
まず見るべき点は式の形が左辺と似ている所です。誘導が使えそうですよね。

誘導を上手く使うコツですが、「誘導の部分と問題文の該当部分の違いを上手く見分けること」です。今回であればnがp-1に変わっています。また、(1)でnは"正の偶数"でしたが、p-1は"素数-1"ですよね。

ここの違いは何かあるでしょうか？？

まず整数問題で素数が出たら、「2とそれ以外」という見方をするのは演習量をこなせば分かってきます。素数の中でも2だけ偶数で稀有、と認識できていればOKです。(ここは基本問題的な解法暗記の部分)

素数-1は、素数が2のときだけ奇数、素数が2以外のときは偶数になりますよね！

ですので、2か2じゃない素数かで分けます。2じゃない素数のときは(1)の条件と一致するので使えそうですよね。まずは使いましょう！

○pが2以外の素数のとき
(1)より左辺は3の倍数です。ということは右辺も3の倍数になります。p^k、つまり素数の累乗が3の倍数ということはpは3以外ありえないですよね。ここは素数ならではです。
ですのでp=3から左辺に代入するとk=1と決まります。

○pが2のとき
代入していくとk=0になりますね。

以上から(p,k)=(3,1),(2,0)となりました！

このように、「基本問題の解法はすぐに出ておくようにする」「誘導から常に考えていく(誘導と問題文の違いを認識し、見分けていく)」ことの重要性がわかったと思います。また、基本問題というのは、教科書や青チャートにある典型問題もそうですが、素数は2とそれ以外に分ける、といったような"応用問題でよく出てくるテクニック"もそうです！これは演習量を詰まないと中々インプットされないので、「演習量が大切」なのも再認識できるでしょう。

また、1問に時間をかけて思考していくこともとても大切です！最終的にその標準問題の解き方を覚えられると役には立ちますが、思考力というのは思考する時間を取らないと中々伸びません。1問に10分は考える時間を取りましょう！

めちゃくちゃ長くなって申し訳ないですが、参考になれば幸いです！！

1d:Tc1f,　こういった問題独自の定義は、だいたい文字を含んでいることが多いです。例えば、
・「nを正の整数とし、3^nを10で割った余りをanとする。」（東京大2016文系）
・「正の整数nの各位の数の和をS(n)で表す。」（一橋大2018）
・「nを2以上の整数とする。金貨と銀貨を含むn枚の硬貨を同時に投げ、裏が出た金貨は取り去り、取り去った金貨と同じ枚数の銀貨を加えるという試行の繰り返しを考える。初めはn枚すべてが金貨であり、n枚すべてが銀貨になった後も試行を繰り返す。k回目の試行の直後に、n枚の硬貨の中に金貨がj枚だけ残る確率をPk(j)（0≦j≦n）で表す。」（東北大2019文系）
のように。あなたが挙げて下さった例でもそうですね。
　ご存知のように、数学で文字が使われるのはそこに入る値が不特定であるときなので、逆にいえば、自分で具体的な値を代入して実験してみれば良いわけです。k-連続和でいえば、m=1、k=2とすると、3＝1+2という等式になり、3は2-連続和であることになります（相談文のk+1はおそらくkー1の間違いですね。でなければ、nはk+2個の連続する自然数の和になってしまうので）。ちゃんと、n(3)がk(2)個の連続する自然数(1→2)の和であるという定義に則ってますね。2019年文系の確率も、例えばk＝1を代入してみると、P1(j)は「n枚の金貨を同時に投げ、そのうちj枚が表で他が裏になる確率」のことを言っているのだとわかります（ちなみにこれは小問⑴）。反復試行の確率を考えればすぐ解けますね。すると、次はk＝2、その次はk＝3、と実験数をどんどん増やしていけば、Pk(j)の内容もいずれわかるはずです。試行の手順上、残るj枚は必ず全ての試行において表でなければならず、他方それ以外の金貨はすべて、k回のうちのどこかで裏が出ればいい（全て表で残る場合の余事象）わけですから、「n枚の金貨のうち、k回の試行の直後に残るべきj枚はk回とも全て表が出て、それ以外のn−j枚はk回の試行で少なくとも一回裏が出る確率」とわかります。ここまで日本語として簡略化できれば、Pk(j)（特に、k≧2）の値もそこまで苦戦せずに出せそうですね（ちなみにこれは小問⑵）。
　このように、なるべく簡単な値から代入して実験を繰り返すことで、独自の定義が何を言っているのかは帰納的に理解できることが多いです。文字が多かったり、分かりにくい表現だったりして、複雑で難しく感じる定義が出てきたら、まずは実験してみることを心がけると良いと思います。文系の問題ですが、もしまだ解いてない場合はネタバレになってしまい申し訳ございません。2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=BQIkDhojwTnUwW4wIkBj","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"aとbの求め方が分からない"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",3,") コメント(",0,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"10/12 13:10"}]]}],["$","div",null,{"className":"coach-mark mb-4","children":"UniLink利用者の80%以上は、難関大学を志望する受験生です。これまでのデータから、偏差値の高いユーザーほど毎日UniLinkアプリを起動することが分かっています。"}],["$","div",null,{"className":"mb-4","children":["$","$L13",null,{"clientImageUrl":null,"clientUserName":"はやて","infoString":"高3 兵庫県 兵庫大学現代ビジネス学部(38)志望","adviceId":"BQIkDhojwTnUwW4wIkBj"}]}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap","children":[["$","div","consultation-part-0",{"children":[null,"関数f（ｘ）＝X3乗＋aX2乗＋bX＋1が、X＝-1で\n極大値をX＝3で極小値をとるように、\n定数aとbの値を定めよ。また、極値を求めよ"]}]]}],["$","div",null,{"className":"pt-4","children":["$","$L14",null,{}]}],null]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"回答"}],["$","div",null,{"className":"mb-4","children":["$","$L15",null,{"adviserImageUrl":null,"adviserName":"こばり","adviserDepartment":"東京大学文科二類","adviceId":"BQIkDhojwTnUwW4wIkBj"}]}],["$","div",null,{"className":"coach-mark 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":[["$","div","advice-part-0",{"children":[null,"f(x)=x^3+ax^2+bx+1\nf’(x)=3x^2+2ax+b=0とすると\nx=(-a±√a^2-3b)/3=-1,3なので\nこの連立方程式を解いてa,bを求め、元のf(x)に代入した後にx=-1,3を代入すれば極値も出ます。"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":null,"adviserName":"こばり","adviserDepartment":"東京大学文科二類","adviceId":"BQIkDhojwTnUwW4wIkBj","numberOfFan":3,"clipsAvg":4.625,"adviceRateAvg":4.3076923076923075,"profile":""}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink パンフレットバナー画像","className":"mt-4 rounded"}]}]}],["$","div",null,{"className":"pt-4","children":["$","$L17",null,{"id":"adsbygoogle-init-under-advice"}]}]]}],["$","div",null,{"className":"flex justify-between","children":[["$","h1",null,{"className":"text-xl font-semibold","children":["コメント(",0,")"]}],["$","$L18",null,{"adviceId":"BQIkDhojwTnUwW4wIkBj"}]]}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"text-xs p-4","children":"コメントで回答者に感謝を伝えましょう！相談者以外も投稿できます。"}]}],["$","h1",null,{"className":"text-xl font-semibold","children":"よく一緒に読まれている人気の回答"}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"divide-y","children":[["$","div",null,{"children":["$","$L7",null,{"href":"/advice/bU4qMXkBTqPwDZPuncb2","children":["$","div",null,{"className":"flex items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"隣接3項間漸化式"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは、名古屋大学医学部医学科のメイメイといいます。\n(an-an-1)=bnとするとb1は求められないですね。\n\n(an+1)-(an)=2［(an)-(an-1)］\nが出てきているはずですが、\n\nn-1の項があり基本的にn≧2で考えています。\nこれをn≧1に直してみると\n(an+2)-(an+1)=2［(an+1)-(an)］\nとなります。\n単純にnの部分を1ずつずらしただけです。\n\nこの状態で(an+1)-(an)=bn\nと置いてみましょう。\n\nb1が求められるはずです。(ちなみにb2は必要ないです。)\n\nつまり(bn+1)=2(bn)、b1=(a2)-(a1)=8の等比数列に帰着しますね。\n\nこれを解くと、bn=8・2^n-1=2^n+2となります。(2^n-1は2のn-1乗という意味です。)\n\nすなわち、(an+1)-(an)=2^n+2\n\n両辺を2^n+1で割ると\n\n<(an+1)/2^n+1>-(1/2)<(an)/2^n>=2\n\nとなります。\n\n(an)/2^nをcnとすると、(cn+1)=(1/2)(cn)+2\n\nこれを変形して、(cn+1)-4=(1/2)<(cn)-4>\n\nつまり(cn)-4=(-7/2)・(1/2)^n-1=(-7)・(1/2)^n\n\nよってcn=4-7・(1/2)^n\n\nこの両辺に2^nをかけてan=4・2^n-7 (n≧1)\n\nとなります。\n分かりにくくてすいません！\n\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":[]}],["$","path","1",{"d":"M12 6c1.1 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"指数関数を解くコツは"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは！\nこうしんと申します！\n\n指数関数…というと範囲が難しいので、\n最大最小問題の解き方→指数関数の処理方法\nという形で話を進めていきますね！\n\nまず最大最小問題ですが、これは方程式・関数を扱う分野で出てきます。\nこの分野の攻略方法は以下の通りです\n・文字を見分ける\n・解答法を知る\n（方程式として解く、関数として解く、不等式として解く）\n\n一つずつ説明していきますね。\n\n・文字を見分ける\n文字は、定数と変数があります。物理ではこれがはっきり決まってますが、数学では全く別の性質で、定数でさえ値を動かすことがあります。\nなので\n定数…中心にはない文字\n変数…中心に扱っていく文字（〜と解く、微分する、といった文字の中心となります）\nこれをまず見分ける必要があります。\n見分け方は、定数が「分布（どういう値をとるのか？）を知りたい文字」であるという性質がある点です。他には、定数の方が次元が高い、扱いづらいという特徴がありますね。\nこうして、変数を絞り込んでおきます。\n変数は１個にしてください。\n\n・解答法を知る\n解答法は３つに分かれます。\n\n方程式としてみる\n→解の配置（０より大小となる点を探す）・座標・対称式\n関数としてみる\n→微分してグラフを描く\n不等式としてみる\n→実数の２乗は０以上を使う、コーシーシュワルツ、相加相乗平均\n（不等式は難しいので、関数としてみた方が早いです）\n\nこれらの解答法を調べてみてください！完璧にすると対応ができます！\n\n\n最大値というのは、\n・関数がそれ以上に増えない値\n・それを満たすxが一つ定義域に存在する値\nであるという性質を持ちます。\n最小値は、反転した性質ですね。\n\nそのため最大値の候補は絞られます\n→①極大値　②区間の端\nこの2点を調べてみましょう。（最小値は反転です）\n\n最後に、最大最小を論じる際に、よく出てくる言葉があるので、それを押さえておきましょう。\n・領域→「接する時」「端の時」に最大最小\n・接する→最短距離があります、注意です\n\nポイントはこんな感じです！\nよく分かんないかもしれませんが、演習しながら見てください！意味がわかってくるはずです！\n頑張ってください！応援してます！"}],["$","div",null,{"className":"flex 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