奥数问题大家讨论

davidbin

oliver_871

这题好像没有什么简单的做法。上面这个流程图可能比较适合小学生理解。
1.数字1显然只有一个位置，所以有图P1
2.从P1出发找数字2的位置，有P2,P3两种可能
3.P3和P2对称，我们只要知道P2有几种可能，就知道P3了，所以P3可以不用继续走下去了，用红色标注。
4.余下步骤类似。利用对称或相似，我们可以大大减少分支，在手工穷举能力之内
5.到了P17（数字7）,只有一个可能了，所以标注为1在左上方。P18,P19和P17对称或相似，可以抄P17的标注，也是1
6.沿流程反推回来。比如P15生成P18和P19，那么P15=P18+P19，P15可能的数字排列是2。
7.红色的可以抄同层对称或相似的。比如P9抄P6。
8.这样算到P1，就是答案42

静雯

谢谢各位版亲的热情支持与辛勤劳动。

静雯

Another method:

fion玲

急求奥数老师或推荐 谢谢  请联系  或QQ505524402

静雯

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frekiwang

静雯 发表于 6-2-2012 09:52
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1. Eliminatin all zeros, 90! has [90/5]+[90/25]=21 5s in its prime factorisation, so it ends with 21 zeros since we can easily find 21s 2s in its prime fractorisation. Thus, we are interested in finding the last two digits of x=90!/10^21, and the answer is the remainder of x when divided by 100.

2. By looking at the factorisation of x, we can easily conclude it is divisibly by 4.

3. If the remainder of x when divided by 25 is k, the remaindr of x when divided by 100 must be k, k+25, k+50 or k+75.

4. First of all, we need to reserve some numbers to have a total of 21 2s as factors. In 10,20,30..90 there are 1+2+1+3+1+2+1+4+1=16 2s. Here I will reserve 4 and 8 to have addtion 5 2s. After all the 2s and 5s are 'used up', the number left in from 5,10,15...85,90 are 1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 11, 3, 13, 7, 3, 1, 17, 9.

5. Note (5n+1)(5n+2)(5n+3)(5n+4) = 24 (or -1) (mod 25), which means
(11)(12)(13)(14)=-1(mod 25)
(16)(17)(18)(19)=-1(mod 25)
(21)(22)(23)(24)=-1(mod 25)
...
(81)(82)(83)(84)=-1(mod 25)
(86)(87)(88)(89)=-1(mod 25)

So, (11)(12)(13)(14)(16)(17)....(88)(89)=(-1)^16=1(mod 25)

6.
Include the last few numbers that we have not count: 1, 2, 3, 6, 7, 9, which the product (1)(2)(3)(6)(7)(9) = (36)(7)(9) = (11)(7)(9) = (77)(9) = (2)(9)=18 (mod 25)
Include the numbers left from 10, 15...., 90, which the product 1, (3)(3)(7)(9)(11)(3)(13)(7)(3)(17)(9) = (27)(21)(33)(91)(51)(9) = (2)(-4)(8)(-9)(1)(9) = 5184 = 9 (mod 25)

7. so x = (1)(18)(9) = 162 = 12 (mod 25)
8. so the last two digits must be 12, 37, 62 or 87, but only 12 is divisibly by 4.
9. Therefore the last two digits are 12.

OMG, I don't think primary school students can follow, perhaps try to find a pattern?

仙都峰居士

答案应该是64，算法也很复杂，况且小学没有学到mod的算法。

oliver_871

答案是12，frekiwang是正确的。

需要哪种拿数学当玩具的小学生才能理解这个解法。

仙都峰居士

sorry， 太勤劳了，一口气算到100。