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# [M2] Four 5** maths questions [Core & M2]

## Four 5** maths questions [Core & M2]

(1) [Core] Let n be a positive odd number. Prove that (x+y+z)^n - x^n - y^n - z^n is divisible by (x+y)(y+z)(z+x).
(2) [Core] Prove that (x+y+z)^7 - x^7 - y^7 - z^7 is divisible by 2.
(3) [M2] Prove that (x+y+z)^7 - x^7 - y^7 - z^7 is divisible by 7.
(4) [M2] Prove that (x+y+z)^7 - x^7 - y^7 - z^7 is divisible by 42.

Try these 4 challenging qs out! TOP

 風之男 學院師父 發短消息 加為好友 當前離線 2# 大 中 小 發表於 2020-12-29 11:57 PM (第 261 天) 只看該作者 [顯示] [隱藏] Minor nitpick: it's better to specify x, y, z are integers (although it's kind of implied, or else divisibility wouldn't be something worth discussing). Here are some hints for those who are interested. --- SOLUTION SPOILERS --- (1): Think about factor theorem. Suppose f(x) = (x+y+z)^n - x^n - y^n - z^n. How do we show that f(x) is divisible by x+y? (2): From (1), we know that the expression is divisible by (x+y), (y+z) and (x+z). Can they be all odd? (3): At least how I did it, it requires trinomial expansion, which, although it's not covered in DSE M2, is still a nice result to know. (x+y+z)^7 = x^7 + y^7 + z^7 + [other terms. Maybe think about a simpler trinomial first?]. (4): While I haven't actually done the proof, (4) is the same as showing the expression is divisible by 3, then combining with (2) and (3) will give you the desired result. It's not too hard of a proof if you think that every integer can be expressed in either 3k, 3k+1 or 3k-1, where k is an integer, and just plug it in the expression to check (27 cases! However many of them are equivalent to each other), although very clunky (much simpler if you are comfortable with modular arithmetic, a.k.a. contest math's favorite). I wonder if there are any other methods... [ 本帖最後由 風之男 於 2020-12-30 12:42 AM 編輯 ] UID216418 帖子932 精華0 積分115 閱讀權限40 在線時間1610 小時 註冊時間2011-5-4 最後登錄2021-9-15  查看詳細資料 TOP

## 回覆 2# 風之男 的帖子

Good tips! And thanks for reminding me that x, y, and z should be integers at least for (2)-(4).
(1) it’s an advanced problem on the remainder theorem.
(2) it’s about parity checking. (You can also directly check the parity of (x+y+z)^7-x^7-y^7-z^7)
(3) it involves trinomial expansion. (It’s out-c but a student who targets a 5** in M2 should be capable of solving this problem)
(4) it requires the trick mentioned by 風之男 to prove the divisibility by 3. (you will need a calculator if using 3a, 3b+1, 3c+2) and I only examined 8 cases due to symmetry.
A student who aims high should at least know how to tackle (1)-(3) [(3) for M2 students].
FACT FUN: If k divides (x+y+z)^7-x^7-y^7-z^7, then the maximum value of k is 42 (the magic number 42)! I will post the solution tmr. [ 本帖最後由 blazehaze 於 2020-12-30 12:36 AM 編輯 ]

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 blazehaze 資深師父  發短消息 加為好友 當前離線 4# 大 中 小 發表於 2020-12-31 09:01 AM (第 259 天) 只看該作者 solution here 附件: 您所在的用戶組無法下載或查看附件 哀吾生之須臾，羨長江之無窮 UID154480 帖子3459 精華0 積分328 閱讀權限50 在線時間3132 小時 註冊時間2010-3-16 最後登錄2021-9-8  查看詳細資料 TOP
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