『今週の問題』第101回　解答

◆Earl Haig Secondary School (Toronto, Ontario Canada)
　Keigo Kawaji さんからの解答

Q1:

The cards are in the following order:
2 3 1 2 1 3

Q2:

The cards are in the following order:
2 3 4 2 1 3 1 4

Q3:

The cards are in the following order:
7 2 4 6 2 3 5 4 7 3 6 1 5 1

Omake 1:

If we observe the total number of cards for N = 5 and 6, there are respectively 2x5, 2x6 cards each.
The largest card: let's say 5. this would be be placed more than half of the slots apart.
This card would take up 7 slots.
x x x 5 - - - - - 5

Now let us look at the number of slots that is within the cards.
In other words, if N is the # of cards, and N = 5 in this example,
there are 1+2+3+4+5 = 15 slots available.

But the # of the available slots must be even, (eg. 2, 4, 6, 2n) as there are "pairs of cards" (refer to the theorem on parity)

Therefore, since the slots are odd, (eg. 1, 3, 5, 2n-1) there is no possible combination for 5.

Hence, this will not work for 6 as 1+2+3+4+5+6 = 21

Omake 2:

The best way to evaluate is to add the numbers from 1 to the sum, and see if it is even or odd.

If even, then there is a possible combination.

If odd, there is no possible combination, as it denies the rule of parity, thus will not have a satisfying solution.

In details: if N = 4, there are 8 cards.
There are 4 cards supposed to be in between the two 4s, there are 3 cards supposed to be in between the two 3s.
2 cards in between the 2s and a single card between the 1s.

since the sum of all the number up to n is divisible by 2, it will work for 4

２ |

n Σ k=1

k

◆広島県　清川　育男 さんからの解答

【問題１】

２ ３ １ ２ １ ３
３ １ ２ １ ３ ２

【問題２】

２ ３ ４ ２ １ ３ １ ４
４ １ ３ １ ２ ４ ３ ２

【問題３】

2  5  6  2  3  7  4  5  3  6  1  4  1  7
5  2  6  4  2  7  5  3  4  6  1  3  1  7
7  1  3  1  6  4  3  5  7  2  4  6  2  5 
7  1  4  1  6  3  5  4  7  3  2  6  5  2

１  １○１　　　　　　　　　　３個
２  ２○○２　　　　　　　　　４個
  ．．．．．．．
  ．．．．．．．

Ｎ  Ｎ○○・・・○○Ｎ　　（Ｎ＋２）個

1) N=4k
　 (4k+5)*4k/2=2k*(4k+5)

2) N=4k+1
　(4k+6)*(4k+1)/2=(2k+3)*(4k+1)

3) N=4k+2
　(4k+7)*(4k+2)/2=(2k+1)*(4k+7)

4) N=4k+3
　(4k+8)*(4k+3)/2=2(k+2)*(4k+3)

1),4) の場合が偶数となる。

したがって、可能であるための必要条件は、
N≡0 (mod 4) または N≡3 (mod 4)

ゆえにNが
N≡1 (mod 4) または N≡2 (mod 4) のときは不可能となる。

◆愛知県　Y.M.Ojisan さんからの解答

【問題１】

２−３−Ａ−２−Ａ−３

【問題２】

２−３−４−２−Ａ−３−Ａ−４

【問題３】

２−４−７−２−３−６−４−５−３−Ａ−７−Ａ−６−５

【おまけ１，２】

Ｎ mod 4 が　１と２のものはできない。

証明：この問題は過去問の第８９回の問題３の焼き直しである。
不可能の証明は偶奇性によっているので、輪になっているかどうかは関係なく、そのまま適用でき、おなじ結論になる。
（なお、出来る方 N mod 4=[0|3] は[十分条件]である。）

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