Can you solve the sea monster riddle? - Dan Finkel

2,715,046 views ・ 2020-04-02

TED-Ed


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翻译人员: Meng Ren 校对人员: Yanyan Hong
00:06
According to legend, once every thousand years
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传说,每过一千年,
00:10
a host of sea monsters emerges from the depths to demand tribute
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一群海怪就会从海底深处浮出水面,
向漂浮之城阿特兰蒂卡索要上贡。
00:15
from the floating city of Atlantartica.
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00:18
As the ruler of the city, you’d always dismissed the stories…
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作为全城的统治者, 你对传说总是不以为然……
00:22
until today, when 7 Leviathan Lords rose out of the roiling waters
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直至今日,七条利维坦神兽 从滚滚的海浪里翻涌而出
00:28
and surrounded your city.
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包围了你的城邦。
00:30
Each commands 10 giant kraken,
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每条神兽指挥着十只北海巨怪,
00:33
and each kraken is accompanied by 12 mermites.
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每只北海巨怪带领着十二只魔蟹。
00:37
Your city’s puny army is hopelessly outmatched.
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而你城里的兵力远远不及。
00:41
You think back to the legends.
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你回想起那个传说。
00:43
In the stories, the ruler of the city saved his people
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故事中提到, 城邦的统治者为了拯救子民,
00:46
by feeding the creatures a ransom of pearls.
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向怪物们上贡珍珠作为赎金。
00:50
The pearls would be split equally between the leviathans lords.
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珍珠必须能被 所有利维坦神兽均分。
00:54
Each leviathan would then divide its share into 11 equal piles, keeping one,
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而每条神兽将它所得的 分为十一等份,自留一份,
01:00
and giving the other 10 to their kraken commanders.
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其余的分给十只北海巨怪。
01:04
Each kraken would then divide its share into 13 equal piles,
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每只北海巨怪又将 它的所得分为十三等份,
01:09
keeping one, and distributing the other twelve to their mermite minions.
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自留一份, 其余的分给十二只蟹卒。
01:14
If any one of these divisions left an unequal pile or leftover pearl,
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若是任意一次分配不能被均分 或最后珍珠有剩余,
01:20
the monsters would pull everyone to the bottom of the sea.
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海怪们就会将整个城邦拖入海底。
01:24
Such was the fate of your fabled sister city.
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传说中你的邻城就惨遭其毒手。
01:28
You rush to the ancient treasure room and find five chests,
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你赶紧跑去藏宝阁, 找到了五只大宝箱,
01:32
each containing a precisely counted number of pearls
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每个箱子里装有特定数额的珍珠,
01:35
prepared by your ancestors for exactly this purpose.
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这正是祖先们为今天准备的。
01:40
Each of the chests bears a number telling how many pearls it contains.
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每个藏宝箱的铭牌上都刻着数字, 代表所装的珍珠数量。
01:44
Unfortunately, the symbols they used to write digits 1,000 years ago
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但不巧的是,祖先的数字 是一千年前的古文字,
01:49
have changed with time,
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如今时过境迁,
01:51
and you don’t know how to read the ancient numbers.
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而你不认得这些古老的数字。
01:54
With hundreds of thousands of pearls in each chest, there’s no time to recount.
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箱子里的珍珠成百上千, 根本没有再次清点的时间。
02:00
One of these chests will save your city
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它们其中一只将会拯救你的城邦,
02:03
and the rest will lead to its certain doom.
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而其余的则无力回天。
02:06
Which do you choose?
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你要如何选择呢?
02:07
Pause the video to figure it out yourself.
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[ 若要自己找到答案,就暂停一下视频 ]
02:09
Answer in 3
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[ 答案 3 秒后揭晓 ]
02:10
Answer in 2
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[ 答案 2 秒后揭晓 ]
02:12
Answer in 1
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[ 答案 1 秒后揭晓 ]
02:15
There isn’t enough information to decode the ancient Atlantartican numeral system.
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虽然没有足够的信息去揭秘 古阿特兰蒂卡的计数系统。
02:20
But all hope is not lost,
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但还是有一线希望,
02:22
because there’s another piece of information those symbols contain:
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因为那些符号包含着另一类信息:
02:26
patterns.
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规律。
02:27
If we can find a matching pattern in arabic numerals,
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如果我们能找到 相对应阿拉伯数字的规律,
02:31
we can still pick the right chest.
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就能选出正确的箱子。
02:34
Let’s take stock of what we know.
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我们先来总结一下已知的信息。
02:36
A quantity of pearls that can appease the sea monsters
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能平息海怪的珍珠数量
02:39
must be divisible by 7, 11, and 13.
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必须能被 7、11、和 13 整除。
02:43
Rather than trying out numbers at random,
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与其随机地试不同的数字,
02:45
let’s examine ones that have this property
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不如来研究一下 这些数字的特性,
02:48
and see if there are any patterns that unite them.
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并找到它们共有的模式。
02:51
Being divisible by 7, 11, and 13
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能被 7、11、和 13 整除,
02:54
means that our number must be a multiple of 7, 11, and 13.
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说明这个数字必须 同时是 7、11、和 13 的倍数,
02:59
Those three numbers are all prime, so multiplying them together
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这三个数字均为质数, 所以将它们相乘
03:04
will give us their least common multiple: 1001.
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就得到了最小公倍数:1001。
03:08
That’s a useful starting place
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这个信息非常重要,
03:10
because we now know that any viable offering to the sea monsters
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现在我们知道任何一个 能满足海怪们要求的数字
03:14
must be a multiple of 1001.
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必须是 1001 的倍数。
03:17
Let’s try multiplying it by a three digit number,
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让我们来试一试 用它乘以一个三位数,
03:20
just to get a feel for what we might get.
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看看能得到怎样的结果。
03:22
If we try 861 times 1001, we get 861,861,
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用 861 乘以 1001, 我们得到了 861861,
03:30
and we see something similar with other examples.
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我们找到了所有数组 共有的规律。
03:33
It’s a peculiar pattern.
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它们都有一个特定模式。
03:35
Why would multiplying a three-digit number by 1001
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为什么用任意一个三位数乘以 1001,
03:39
end up giving you two copies of that number,
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得到的结果从左到右读就是该三位数
03:42
written one after the other?
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重复了两次?
03:43
Breaking down the multiplication problem can give us the answer.
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分解这个乘式就能找到答案。
03:47
1001 times any number x is equal to 1000x + x.
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1001 乘以未知数 X 等于 1000 X + X 。
03:53
For example, 725 times 1000 is 725,000, and 725 x 1 is 725.
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例如,725 乘 1000 是 725000 , 725 乘 1 是 725,
04:03
So 725 x 1001 will be the sum of those two numbers: 725,725.
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所以,725 乘 1001 就是 两个数字总和:725725。
04:12
And there’s nothing special about 725.
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我们可以把 725 换成任意一个三位数。
04:15
Pick any three-digit number,
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随机选择一个三位数,
04:17
and your final product will have that many thousands, plus one more.
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你得到的结果将会是 它的一千倍与它的一倍的和。
04:22
Even though you don’t know how to read the numbers on the chests,
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因此,即使你看不懂 宝箱上的古数字,
04:26
you can read which pattern of digits represents a number divisible by 1001.
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你仍然能找出符合 1001 整除规律的那一组。
04:32
As with many problems, trying concrete examples can give you an intuition
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许多问题是同一个道理。 面对抽象而神秘的表象时,
套入一个具体的例子, 能帮助我们洞察事物背后的规律。
04:37
for behavior that may at first look abstract and mysterious.
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04:42
The monsters accept your ransom and swim back down to the depths
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最后,海怪们接受了赎金, 潜回了海底深处,
04:46
for another thousand years.
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等待下一个千年轮回。
04:48
With the proper planning,
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经过周全地计划,
04:49
that should give you plenty of time to prepare for their inevitable return.
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你将会有充裕的准备时间 迎接海怪们下一次不可避免的回归。
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