一道小数学题

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原作者：InCaseYouLikeIt www.ddhw.com

有n个数：A₁, A₂, A₃, A₄,,,,,,,A_n

 

这些数要么是1，要么是 -1

 

如果 A₁A₂A₃A₄ + A₂A₃A₄A₅ + ..... + A_n-2A_n-2A_nA₁ + A_n-1A_nA₁A₂ + A_nA₁A₂A₃ =0

 

请证明 n 是 4的倍数。

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孜孜不倦@.

  idunknow @.@

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idiot94

  

野 菜 花

Why is the sum of the 4 products containing d equal to 0? abcd+bcde+cdef+defg=0? www.ddhw.com

野 菜 花

每个数在和中出现4次, 如果将一个(-1)改成(+1), 这4个积都改变符号, 总和会相差4的倍数.(具体见下面) 如果把所有的(-1)改成(+1), 总和还是4的倍数.另一方面和里共有n个乘积1,所以和是n. 所以n是4的倍数.   若这4个乘积都是+1, 就变成4个-1, 新和减少8,=-8. 若这4个乘积是三正一负, 就变成三负一正, 新和减少4, =-4. 若这4个乘积是二正二负, 就变成二负二正, 新和不变, =0. 若这4个乘积是一正三负, 就变成一负三正, 新和增加4, =4. 若这4个乘积都是-1, 就变成4个+1, 新和增加8,=8.

孜孜不倦@.

assume Ai is one of term in the sequence. |x x x Ai| + | x x Ai x| + | X Ai X X| + | Ai X X X| + ....... (X is 1 or -1) Ai will always appear 4 times in the sequence. we can rewrite as a b c (d =Ai) e f g . ... a b c d b c d e c d e f d e f g Ai=1/-1, => a(bc) + (b(c)e) + (c(e)f) + (ef)g = 0www.ddhw.com case1 if abc + bce = 0, a = -e if cef + efg = 0, c = -g case2 if abc + cef = 0, ab = -ef if bce + efg = 0, bc = -gf case 3 if abc + efg = 0, abc=-efg if bce + cef = 0, b=-f from case 1 and 3, a = -e, b=-f, c=-g, but it can't apply to case 2. if case 1 is true and case 3 is true, then abc + cef = -1*(bce + efg) as the unique solution. we can rewrite the sequence again X,Y,Z,Ai,-X,-Y,-Z,-Ai....www.ddhw.com and xyzAi + yzAi-x +Zai-x-y, + Ai-x-y-z = 0   www.ddhw.org---