【41】“有一牧场,已知养牛27头,6天把草吃尽;养牛23头,9天把草吃尽。如果养牛21头,那么几天能把牧场上的草吃尽呢?并且牧场上的草是不断生长的。” 到牛死了,也吃不完。 ------------------------------------ 0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;">这题出过,跳过。 ---------------------------------- |
除了42外,41、47也都在这坛上讨论过。 41中应该明确草长的一些约定。 46好象也应修改。 类似于49的问题,小时候(4~5岁吧)玩过,叫“抢三十”。 |
41、42、47、48都已经讨论过。46应该修改,50给的条件不充分,重物的钢性(包着棉花或泡沫塑料跟光光的铁球石块可不一样)、山羊脑门毛皮和肉的柔性都不知道,重物的形状也有关呀(象狼牙棒那样的重物会比较容易砸破脑门的)。 又到周末了。对其余的四个,俺来答答试试。 【43】已知: 每个飞机只有一个油箱, 飞机之间可以相互加油(注意是相互,没有加油机) 一箱油可供一架飞机绕地球飞半圈,问题:为使至少一架飞机绕地球一圈回到起飞时的飞机场,至少需要出动几架飞机?(所有飞机从同一机场起飞,而且必须安全返回机场,不允许中途降落,中间没有飞机场)
【44】有十瓶药,每瓶里都装有100片药,其中有八瓶里的药每片重10克,另有两瓶里的药每片重9克。用一个蛮精确的小秤,只称一次,如何找出份量较轻的那两个药瓶? 0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;">
【45】一个经理有三个女儿, 三个女儿的年龄加起来等于13,三个女儿的年龄乘起来等于经理自己的年龄,有一个下属已知道经理的年龄,但仍不能确定经理三个女儿的年龄,这时经理说只有一个女儿的头发是黑的,然后这个下属就知道了经理三个女儿的年龄。请问三个女儿的年龄分别是多少?为什么? 2,2,9。因为在青壮年年龄范围内,只有36=6x6x1=2x2x9满足要求。
【49】假设排列着100个乒乓球,由两个人轮流拿球装入口袋,能拿到第100个乒乓球的人为胜利者。条件是:每次拿球者至少要拿1个,但最多不能超过5个,问:如果你是最先拿球的人,你该拿几个?以后怎么拿就能保证你能得到第100个乒乓球? 0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;"> |
42) 667 根胡萝卜 Take the first 1000根胡萝卜, stop at 333公里, leave 334根胡萝卜; go back, take another 1000根胡萝卜, stop at 667公里 and leave 333根胡萝卜(along the way pick up the 334根胡萝卜 that were left at 333公里); go back, take the last 1000根胡萝卜 (along the way pick up the 333根胡萝卜 that were left at 667公里). 44) Mark 十瓶药 as A,B,C,-----,J From each bottle take a different number of pills, making sure that the sum of any two of these numbers does not equal the number of pills selected from any bottle. Example: take 1 from A, 2 from B, 4 from C, and 7,10,13,16,19,21, and 27 from the remaining bottles. 称一次; you should be able to 找出份量较轻的那两个药瓶 by the difference in weight from 1200克. 0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;"> |
俺44题的解答错啦,每瓶才100片药,2i 会超过100。正确的答案之一是0, 1, 2, 4, 7, 12, 20, 33, 54, 88。答案不唯一。 |
If the weight is 1160 (or 1163, 1166, 1169, 1171, 1172, 1174, 1177, 1180, 1183, 1186, 1189) g, what is your conclusion? |
I'm so sorry for missing your post; I just noticed your reply to my post. Here is my answer to your question. When the sum is 1160, then,份量较轻的那两个药瓶 are H and I 1163 ---- E and G 1166 ---- B and I 1169 ---- E,I 1171 ---- E,H 1172 ---- D,I 1174 ---- D,H 1177 ---- E,G 1180 ---- D,F 1183 ---- D,E 1186 ---- A,F 1189 ---- A,E |
49】假设排列着100个乒乓球,由两个人轮流拿球装入口袋,能拿到第100个乒乓球的人为胜利者。条件是:每次拿球者至少要拿1个,但最多不能超过5个,问:如果你是最先拿球的人,你该拿几个?以后怎么拿就能保证你能得到第100个乒乓球? 0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;"> Answer: 如果是最先拿球的人,该拿4个;以后拿 factor of 5 就能保证你能得到第100个乒乓球. |
例如,当总和是1160时,为什么答案不是F和J而一定是您所说的H和I呢? 其余类似。 |
"factor of 5" 是什么意思? |
Oops! My mistake. Below is my corrected answer. Hope that there will not be any loose ends this time. 44) Mark 十瓶药 as A,B,C,-----,J From each bottle take a different number of pills, making sure that the sum of any two of these numbers does not equal the sum of any other two. Example: take 1 from A, 2 from B, 3 from C, and 6,10,18,30,43,54, and 60 from the remaining bottles. 称一次; you should be able to 找出份量较轻的那两个药瓶 by the difference in weight from 2270克. |
A factor of five means a whole number that is evenly divisable by five (no remainder). |
43+18=60+1 |
好吧,咱们来玩一把。 您先拿了4个,俺拿2个。您再拿4个以保持剩下的是5的倍数,对吗? |
No, 俺拿3个。 |
那就剩给俺91个。俺拿1个,给您剩90个,您拿吧。 |
91- 4 not = factor of 5; |
您怎么不玩下去了?剩给您90个,该您拿了。您再拿几个呀? 俺告诉您,剩给您90个,您就输定啦。 |
Ha,Ha,I don't think I'm lost yet. The number can't be 91 or 9o; neither one minus 4 is the factor of 5. Below is my answer to the question: Answer: 如果是最先拿球的人,该先拿4个;以后拿 factor of 5 就能保证你能得到第100个乒乓球. |
Using my strategy, the number can't be 91 or 90. But if you insist on playing 90, I'll take away 3; if 91, I'll take away 2. |
看清楚原题中“条件是:每次拿球者至少要拿1个,但最多不能超过5个”。 俺留给您90个,您就拿3个,对吗?这样,俺就再拿3个,留给您84个。 |
留90个??? (I'm afraid it is going to take long time to finish!) I thought you meant that we already took away 90 or 91. Anyway, this time I'll take away 3. (Total taken away is 19 at this moment.) Can we think of a better way to play? I'm afraid that if I keep playing this way, I will be worn out pretty soon. Before reaching the 80s (I mean before taking away 80); you can play my part for me by fulfilling the "factor of 5" method. For example: next, if you take 1, I will take 4; If you take 3, I'll take 2;---- If you take 5, I'll take 5,---- etc. Afterward, you can post all your procedures before you reach 80. Agreed? |
好吧,不逗您了,就跟您直说。 第一步拿4个是唯一正确的取法。留给对方96个,这是6的倍数。这就已经胜卷在握,以后只要保持以6的倍数留给对方,也就是当对方拿X个,自己就相应地拿6- X个就行。正因为知道您没有掌握这一规律,俺就拿2个来试试,这时您正确的应对应该是拿4个,以保持留下6的倍数。可惜您放弃了获胜的机会,回答“No, 俺拿3个”。这样,俺当然就只拿1个,抢占90,并宣称“您就输定啦”。这里,6=5+1。您那个“factor of 5”是错误的。明白了吧。 俺小时候就和小伙伴们玩过类似于这个题的游戏“抢30”。 |
You won !!!
I admit that my original method was wrong.
The new method is: Take away 4 at first. Then make sure that each time the number of ping-pong balls that you and your opponent take away totals 6. |
I admit that my original method was wrong.
The new method is: Take away 4 balls at first. Then make sure that each time the number of ping-pong balls that you and your opponent take away totals 6.
Thanks to Cool-eye Lili. Playing the game with her makes me think more thoroughly. |
别客气! |
12天 |
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