Shall we?
Let's take a look at this multiplication table
25 as the Tail
As you can see above, any number which ends with 5 if squared will have 25 as the tail.
Prefix — Yellow Colored Number
I put yellow color for the opening number sequence because it is interesting.
Let's see the 15 x 15 line.
If we isolate the first number, which is 1, and we do:
1 x (1 + 1) = 2It yields 2, the opening of 15² result, 225.
Another example 45 x 45 or 45²:
We take the first number, which is 4, and we do the same technique
4 x (4 + 1) = 2020 is the starting sequence (before the ending, 25) of the result, 2025.
Last example, the 5005 x 5005 or 5005².
500 x (500 + 1) = 250500
Which is also the first sequence of the result, 25050025.
Speed Squaring
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From the examples above, we can generalise the pattern for the prefix (opening) number sequence as this:
numberBefore5 x ( numberBefore5 + 1 )
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The suffix (tail) will always be 25.
We can attach it as the tail (suffix) of the result.
This is how you speed square for the streets, mate. π
Assuming... the street... only consists of the squares of that odd-5.
Odd-5: 15, 25, 35, 45, 55, ...
15 = 5 × 3
25 = 5 × 5
35 = 5 × 7
45 = 5 × 9
55 = 5 × 11
...
3, 5, 7, 9, 11, and so on are odds.
(2n - 1) in posh. Oi, wiv bleedin' n bigger than one, innit — in Cockney.
In a conversation.
Yo, sup, dude.
25!
Wut? You high?
Blimey! 13225, then!
π€¦
We don't know what the conversation is about — like we listen to the telly waiting for Sadako to crawl out while we're clicking the mute/unmute button. Of a radio. Well, that's unexpected. Can we toss something to that Sadako-gate? π€ Once the inter-blimey-gate is opened, she crawls out, aye? Thus perhaps we can throw something into the telly. π€
π½️ Sadako Yamamura is that dubious specter from the film "Ring" (1998). Because Sadako can bloody identify a phone number by retina match, then call or SMS the target number. Alright, it could happen, but... π€·π€¦ Let's see it from the jolly-world view. π
Call scenario:
VHS πΌ
π§π¦± (Watching VHS.)
π» (Sadako-subroutines are invoked. Daemon process.)
π (Ring, ring.)
π§π¦± Yellow?
π» Nanoka-go ni shinu.
π§π¦± Oi, bazooka go boom boom.
π» (Hangs up.)
π§π¦± Yellow? Blast it, I forgot the pineapple topping.
SMS scenario:
VHS πΌ
π§π¦± (Watching VHS.)
π» (Sadako-subroutines are invoked. Daemon process.)
✉️ (Beep, plink.)
π§π¦± π (Reading SMS.)
✉️ 7-Kakanda yo.
π§π¦± π€³ (Typing reply.)
π§π¦± π¨ (Sends reply.) Yo. (Unsent. 0 balance.)
By those scenarios above, Sadako can always differentiate PSTN and mobile number.
Thus the film was called that. Ring. Because landline phones go "ring ring", not "ketchup mustard".
Hence "ringtone", not "condiment-tone".
Watching that film was like I joined a group of people in the middle of their conversation — no background, build up, just them throwing bottles and spraying ketchup to each other. Which is good for a short attention span, twitchy gremlin as myself. πΊ (Tengu is conjured.)
Next sprint, Sadako supports WhatsApp and Stripe APIs. The developer team calls that payment flow as tap-to-pay death. The dialogue will have a close button, but it will actually double the output. It will also include the bug fix for identifying a human, owning a phone — sometimes, when a chimp saw the VHS or Bob with no phone watched it, the system froze — either Bob watched the chimp or the VHS. Declan forgot to put condition for the null check. High priority on the board, red bold, that.
Usage
Let's find the result of 75².
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The prefix sequence:
- The suffix is 25.
Then the result of 75² is 5625.
Decimal Point
Let's use 0.75.
Calculate the result of 0.75²!
-
Take the 7. So the prefix number sequence:
- The suffix is always 25.
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Last step is to count how many digits after the decimal point.
We have 0.75 (two digits) squared. Then we will have 4 digits behind the decimal point.
Result of 0.75² is 0.5625.
Thank you for visiting, see you again. I hope this bit is useful. π
Hi Johan,
ReplyDeleteThat's some very keen thinking and a neat trick you have there! There's also another pattern: each number that we're squaring is an odd multiple of 5 (that is, 5*1, 5*3, 5*5,...), which we can write as (5*(2n-1))^2. Therefore, the prefactor must depend on which n we choose. If we pair up the first few n's with the prefactors, we find 1:0, 2:2, 3:6, 4:12, 5:20, etc. The relationship is that each n is multiplied by (n-1) (ex: 1*(1-1) = 0, 2*(2-1) = 2, 3*(3-1) = 6). Of course, the prefactors are multiples of 100, so we multiply by 100. Altogether we find that (5*(2n-1))^2 = 100*n*(n-1) + 25, which we could simplify to (2n-1)^2 = 4*n*(n-1) + 1.
Best,
Another Math Wizard
Wow that's really structured. I dig it!
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