【有奖解答】印度式乘法口诀的由来 by ericet

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steempress · @ericet · Dec 25 '18 (edited)

$4.24

【有奖解答】印度式乘法口诀的由来

<center><img src="https://cdn.pixabay.com/photo/2016/11/29/01/16/abacus-1866497_960_720.jpg" alt="" /><br/>
(Image source: <a href="https://cdn.pixabay.com/photo/2016/11/29/01/16/abacus-1866497_960_720.jpg">pixabay</a>)
</center>

之前分享了<a href="https://steemit.com/@ericet/tlu4prnaxp/#@steem-ua/re-tlu4prnaxp-20181224t225245z">印度式乘法口诀</a>，按照口诀的步骤我们可以很快的算出19x19以内的乘法。

前文回顾：

14    x  18 = ？
(被乘数) （乘数）

<ul>
<li>第一步：把被乘数跟乘数的个位数相加 14+8=22</li>
<li>第二步：把第一步的答案乘以10 22x10=220</li>
<li>第三步：把乘数的个位数乘以被乘数的个位数 8x4=32</li>
<li>第四步：被第二步和第三步的答案相加 （14+8）x10+(8x4)=252</li>
</ul>

大家有没有想过这个乘法口诀是怎么由来的吗？
今天我们用小学数学数到的内容来推算出这套口诀。

我们把19x19以内的按照数学公式来显示：
<img src="http://ericet.vornix.blog/wp-content/uploads/2018/12/W@KTCW65TIEJICA3LHC1S.png" alt="" /><br/>

从这套公式里面推算出来的结果就是上面说的印度式乘法的口诀。
不相信吗？我们一步步拆开来看。

<img src="http://ericet.vornix.blog/wp-content/uploads/2018/12/RPD1EA205V@1A0E.png" alt="" /><br/>

其实按照这套公式继续推算，可以推算出99x99以内所有乘法的口诀。

这部分留给大家来推算了
按照我上面推算的公式，推算出<strong>7a x 7b =?</strong> 的口诀，<strong>最快并且附上步骤</strong>的人将获得1 SBI的奖励。

大家快来推算吧~ 这不是送命，是送钱！ <br /><center><hr/><em>Posted from my blog with <a href=&#039;https://wordpress.org/plugins/steempress/&#039;>SteemPress</a> : http://ericet.vornix.blog/2018/12/24/%e3%80%90%e6%9c%89%e5%a5%96%e8%a7%a3%e7%ad%94%e3%80%91%e5%8d%b0%e5%ba%a6%e5%bc%8f%e4%b9%98%e6%b3%95%e5%8f%a3%e8%af%80%e7%9a%84%e7%94%b1%e6%9d%a5/ </em><hr/></center>

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@chloe1 · Dec 25 '18

看到有正确答案了

Posted using [Partiko Android](https://steemit.com/@partiko-android)

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@ericet · Dec 25 '18

下回加油💪

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@chloe1 · Dec 25 '18

70×((70+a)＋b)＋ab

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@ericet · Dec 25 '18

给出过程

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@chloe1 · Dec 25 '18

第一步：把被乘数跟乘数的个位数相加 （70+a）+b
第二步：把第一步的答案乘以 （（70+a）+b）x70
第三步：把乘数的个位数乘以被乘数的个位数 axb
第四步：被第二步和第三步的答案相加 70x（（70+a)+b)+ab

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2 replies

@cnbuddy · Dec 25 '18

吃了吗？还在发愁自己的好文没人发现，收益惨淡吗？记得加上cn-curation标签，让[飞鸽传书](https://steemit.com/@htliao/cn-curation-001-cn) 帮你走出困境吧！假如我的留言打扰到你，请回复“取消”。

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@mia-cc · Dec 25 '18 (edited)

7ax7b=10x（（70+a）+b）+ab
第一步：把被乘数跟乘数的个位数相加 （70+a）+b
第二步：把第一步的答案乘以10 （（70+a）+b）x10
第三步：把乘数的个位数乘以被乘数的个位数  axb
第四步：被第二步和第三步的答案相加  10x（（70+a)+b)+ab

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@ericet · Dec 25 '18

有一步错了，试着把过程展开

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

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@mia-cc · Dec 25 '18

7ax7b=70x（（70+a）+b）+ab
第一步：把被乘数跟乘数的个位数相加 （70+a）+b
第二步：把第一步的答案乘以70 （（70+a）+b）x70
第三步：把乘数的个位数乘以被乘数的个位数  axb
第四步：被第二步和第三步的答案相加  70x（（70+a)+b)+ab

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3 replies

@ericet · Dec 25 '18

比如71x71=5041
按照你的步骤，（71+1）x10+1=721

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

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@steemstem · Dec 25 '18

$0.85

re-ericet-cspm6cjkc2-20181225t172138972z

<div class='text-justify'> <div class='pull-left'> <br /> <center> <img width='125' src='https://i.postimg.cc/9FwhnG3w/steemstem_curie.png'> </center>  <br/> </div> <br /> <br /> 

 This post has been voted on by the **SteemSTEM** curation team and voting trail in collaboration with **@curie**. <br /> 
 If you appreciate the work we are doing then consider [voting](https://www.steemit.com/~witnesses) both projects for witness by selecting [**stem.witness**](https://steemconnect.com/sign/account_witness_vote?approve=1&witness=stem.witness) and [**curie**](https://steemconnect.com/sign/account_witness_vote?approve=1&witness=curie)! <br /> 
For additional information please join us on the [**SteemSTEM discord**]( https://discord.gg/BPARaqn) and to get to know the rest of the community! </div>

👍 steemstem, lafona-miner

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@tydebbie · Dec 25 '18

大腦宣佈當機

Posted using [Partiko Android](https://steemit.com/@partiko-android)

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@wongshiying · Dec 25 '18

1）7a x b
2) (7a x b) x10
3)  a x b
4) [(7a x b) x 10] + ab

是这样吗？

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

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@ericet · Dec 25 '18

第一步错了，是7a+7b

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

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@zerofive · Dec 25 '18

$0.02

7ax7b=(70+a)x(70+b)=4900+70x(a+b)+axb=70x((70+a)+b)+ab

👍 dustbunny, ericet

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@ericet · Dec 25 '18

正确✅

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

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@zerofive · Dec 25 '18

@ericet  按照这样的步骤，算起来好像也不快

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@ericet · Dec 25 '18

是的，只能在19x19内比较快

Posted using [Partiko iOS](https://steemit.com/@partiko-ios)

properties (22)