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After I watched you perform The Monster Mash, I thought about writing a parody called The Monster Maths. Although I tend to think of myself as the undiscovered Weird Al, I just simply couldn't get very far, not beyond... "He did the Maths The Monster Maths It was the classroom smash The Monster Maths..." So I leave this bit for you to do what, if anything, with it.
one of my favorite mental maths tricks (which admittedly is hard to find obvious use cases for out in the wild) is the difference of squares identity. if you want to multiply 17 and 23, just do 20^2 instead, and then subtract the correction. the correction is based on the distance from the midpoint (which is 3), but you have to square it so its really 9. But 20^2 - 9 is easier to my brain than 17 times 23.
it is good for knowing prime couplet pairings by knowing instead all squares like in your example. So from 1 to 100 you don't need to memorise 5150 different pairings of numbers you just memorise the squares from 1 to 100 to multiply all couplets. good trick
I use the same math trick myself. It makes 2 digit multiplication easy for mental work. In fact I used 23x17 as an illustration of this method just the other day to my brother
My favorite trick is using the identity (X + A)(X+B) when A + B = 10, to get that (X + A)(X+B) = X(X + 10) + AB. By choosing A and B to be the last digits of the numbers, AB is completely segregated from X(X + 10) and they can be concatenated. Example: 113 * 117 = (11*12) concatenate with (3*7) = 13221 282 * 288 = (28*29) concatenate with (2*8) = 81216 This makes squaring numbers that end with 5 particularly easy. 115^2 = (11*12) concatenate with 25 = 13225 325^2 = (32*33) concatenate with 25 = 105625
I found out by myself that addition left to right was a lot quicker. I didn't learn this way for subtraction - borrowing was just decreasing the number a bit which I could do easily, but not as quickly as this method. It's nice doing math "wrong" but getting to a result much more quickly.
It's quite flattering seeing a mathematician using the exact same techniques I came up with. I've always thought calculating from left to right was easier
You are great!! man you are really a person who can explain mathematics in an easy and simple way. Btw I want videos on conic sections as I am a little struggling in this topic.So I want videos on it. God bless you!!! Keep going like this..😊😊😊
I sometime use this method too and I don't even remember how I learned it lol. Well it's sometimes easier to subtract from the second number to match the digits of the first number instead of rounding up. For example 8579 - 6681 we can subtract 2 from the second number which gives us 8579 - 6679 which is also easily calculated. Then we subtract 2 from the result because we have subtracted 2 less, giving use 1900 - 2 = 1898. This is easier than 1579 + 319 in my opinion.
The other thing I do is move the numbers to make it easier. 87-39 as an example is 88-40 so that there's an addition first. I also like to factor if it's easily factorable to make a subtraction a multiplication
Nice strat but my aphantasic memory will instantly overflow. Mentally I think I compute either with indistinct abstract ghosts of digits or say things in my head; both are easy to grow into unmanageable mess and strain the brain. Though regular subtraction does it too. IDK I’ll try to apply this when I misguidedly choose to calculate mentally instead of using a calculator.
I do subtraction a really different way, it's an algorithm I "invented" myself. I say it like that because I don't actualy know if anyone else invented it beforehand I haven't found anything. Anyways, I clicked on this video curious to see if my own method was in here but sadly it wasn't. If you're curious on my own method it goes like the following: (it is important to know that a>b in this whole demonstration). a-b=c (this will be our problem). We solve for c using this: b+c=a If you can't do it in your head here's how to do it on paper: First, you write 'b' on the top row, leave a gap and add 'a' on the bottom row, add a line just above 'a' (not directly in the middle) and given the corresponding values of 'b' and 'a' you can solve for 'c' in the middle line. (where you would put the other value). Say you have the number in 'b' be '6' and the number in 'a' be '8'. the number in 'c' would be '2' because 6+2=8. The only problem with the method is if you get '8+c=3'. In which case you write down 5 and carry 10 over. (8+5=13) It works just like addition so in the number after that you do 'b+c+1=a'. And "obviously" if it's the last digit you just add the carry onto the start. - WRONG!! If this happens then you screwed up. You mistakenly made b>a which doesn't work with this setup. womp womp try again. If you desire the negative outcome (b-a=c) first, do it the normal way for the non-negative result (a-b=c) and once complete slap a minus sign at the start of it. There, that's b-a=c. This works because: (a-b) + (b-a) = 0 Therefore, b-a=(-c). It's kind of like 'reverse-addition' so that's what I usually call it, the 'reverse-addition algorithm' alledgedly invented by AliNevada34 although not proven because like I said earlier, I don't know if anyone else invented this beforehand, but it was a method I thought of in my freetime when I was trying to think of a better subtraction algorithm and... well... I use it all the time now. Yes, i have checked all possible permutations of the method, it really does work and i've compared it to the other subtraction methods and everything checks out. I wonder if there's such an algorithm that does division in a reverse-multiplication way, or if there's an absurdly useless way to do addition using reverse-subtraction. That would be so cool having a way to do a division in a format similar to addition, multiplication and subtraction. I don't know if I'll ever find out if I am the true inventor of this, but if anyone has answers about this then let ke know :D.
took about 5 seconds to compute the thumbnail is 1861, though when i force myself to go fast i often make errors, like accidentally getting an answer 200 below what the actual answer is
It's obvious - add 3 to 2597 to get 7400, then add 2000 to get 9400, then add another 58. 3+2000+58 = 2061. The key to getting good at arithmetic is getting comfortable with the value of numbers and then making life easy for yourself in calculating. You could also get the answer easily by 9458 - 7400 is 2058 and then add 3 is 2061. The hard way is to plod through a method by rote, that gives you the right number but, yes, doesn't ever ENGAGE with the values of those numbers. As for 1385 + 3749, add 15 to the first, taje it away from the second and you get 1400 + 3734 which is 5134. Oh and I HATE the way you write 4s.
Hey man, can you help me demonstrate whether lagrange's theorem applies to n-ary groups or not?en.wikipedia.org/wiki/N-ary_group and happy new years btw
How we feeling about a new outro song for 2025? 🤔
Join Wrath of Math to get exclusive videos, lecture notes, and more:
ruclips.net/channel/UCyEKvaxi8mt9FMc62MHcliwjoin
More math chats: ruclips.net/p/PLztBpqftvzxXQDmPmSOwXSU9vOHgty1RO
After I watched you perform The Monster Mash, I thought about writing a parody called The Monster Maths. Although I tend to think of myself as the undiscovered Weird Al, I just simply couldn't get very far, not beyond...
"He did the Maths
The Monster Maths
It was the classroom smash
The Monster Maths..."
So I leave this bit for you to do what, if anything, with it.
one of my favorite mental maths tricks (which admittedly is hard to find obvious use cases for out in the wild) is the difference of squares identity. if you want to multiply 17 and 23, just do 20^2 instead, and then subtract the correction. the correction is based on the distance from the midpoint (which is 3), but you have to square it so its really 9. But 20^2 - 9 is easier to my brain than 17 times 23.
it is good for knowing prime couplet pairings by knowing instead all squares like in your example. So from 1 to 100 you don't need to memorise 5150 different pairings of numbers you just memorise the squares from 1 to 100 to multiply all couplets. good trick
I use the same math trick myself. It makes 2 digit multiplication easy for mental work. In fact I used 23x17 as an illustration of this method just the other day to my brother
My favorite trick is using the identity (X + A)(X+B) when A + B = 10, to get that (X + A)(X+B) = X(X + 10) + AB.
By choosing A and B to be the last digits of the numbers, AB is completely segregated from X(X + 10) and they can be concatenated.
Example: 113 * 117 = (11*12) concatenate with (3*7) = 13221
282 * 288 = (28*29) concatenate with (2*8) = 81216
This makes squaring numbers that end with 5 particularly easy.
115^2 = (11*12) concatenate with 25 = 13225
325^2 = (32*33) concatenate with 25 = 105625
@@julioaurelio Neat!! I knew this thing with squaring numbers ending by 5 but not your generalization!
I found out by myself that addition left to right was a lot quicker. I didn't learn this way for subtraction - borrowing was just decreasing the number a bit which I could do easily, but not as quickly as this method. It's nice doing math "wrong" but getting to a result much more quickly.
A much faster way to subtract (which is an original thought) is to allow negative values. For example, 6772-5919=1(-2)6(-7) or 1060-207=853
Exactly! I thought the same when he said left to right
And here I was expecting 10's complement >.>
Ten's complement is exactly what you do when you round up the second number and have to go back a little bit.
It's quite flattering seeing a mathematician using the exact same techniques I came up with. I've always thought calculating from left to right was easier
I like that you gave us examples to practice with during the video :) Helped me to better engage with the concept rather than just hearing about it.
The distributive law in action.
more like commutative property of addition
Meanwhile I can't keep a single number in my head for more than one second, whatever isn't on paper is gone.
I personally find it significantly less mental workload to round the top number down instead
I find I'm able to do left to right subtraction with borrow, quickly, without needing the round up method.
This reminds me of the "subtract by adding" trick from MinutePhysics!
You are great!! man you are really a person who can explain mathematics in an easy and simple way.
Btw I want videos on conic sections as I am a little struggling in this topic.So I want videos on it.
God bless you!!! Keep going like this..😊😊😊
I do both addition and subtraction right to left
"Are you gonna teach them Urysohn's lemma today?"
"No, The grown ass men will learn how to subract today. They didn't see numbers for a long time."
I sometime use this method too and I don't even remember how I learned it lol.
Well it's sometimes easier to subtract from the second number to match the digits of the first number instead of rounding up.
For example 8579 - 6681 we can subtract 2 from the second number which gives us 8579 - 6679 which is also easily calculated.
Then we subtract 2 from the result because we have subtracted 2 less, giving use 1900 - 2 = 1898. This is easier than 1579 + 319 in my opinion.
The other thing I do is move the numbers to make it easier.
87-39 as an example is 88-40 so that there's an addition first.
I also like to factor if it's easily factorable to make a subtraction a multiplication
Reminds me of that TED Talk on Vedic maths
Nice strat but my aphantasic memory will instantly overflow. Mentally I think I compute either with indistinct abstract ghosts of digits or say things in my head; both are easy to grow into unmanageable mess and strain the brain. Though regular subtraction does it too. IDK I’ll try to apply this when I misguidedly choose to calculate mentally instead of using a calculator.
So proud of myself for figuring out how to do this myself years ago.
Though I thought I alone was the smart one 😔
I do subtraction a really different way, it's an algorithm I "invented" myself.
I say it like that because I don't actualy know if anyone else invented it beforehand I haven't found anything.
Anyways, I clicked on this video curious to see if my own method was in here but sadly it wasn't.
If you're curious on my own method it goes like the following:
(it is important to know that a>b in this whole demonstration).
a-b=c
(this will be our problem).
We solve for c using this:
b+c=a
If you can't do it in your head here's how to do it on paper:
First, you write 'b' on the top row, leave a gap and add 'a' on the bottom row, add a line just above 'a' (not directly in the middle) and given the corresponding values of 'b' and 'a' you can solve for 'c' in the middle line. (where you would put the other value).
Say you have the number in 'b' be '6' and the number in 'a' be '8'.
the number in 'c' would be '2' because 6+2=8.
The only problem with the method is if you get '8+c=3'. In which case you write down 5 and carry 10 over. (8+5=13)
It works just like addition so in the number after that you do 'b+c+1=a'.
And "obviously" if it's the last digit you just add the carry onto the start.
- WRONG!!
If this happens then you screwed up. You mistakenly made b>a which doesn't work with this setup. womp womp try again.
If you desire the negative outcome (b-a=c) first, do it the normal way for the non-negative result (a-b=c) and once complete slap a minus sign at the start of it. There, that's b-a=c.
This works because: (a-b) + (b-a) = 0
Therefore, b-a=(-c).
It's kind of like 'reverse-addition' so that's what I usually call it, the 'reverse-addition algorithm' alledgedly invented by AliNevada34 although not proven because like I said earlier, I don't know if anyone else invented this beforehand, but it was a method I thought of in my freetime when I was trying to think of a better subtraction algorithm and... well... I use it all the time now.
Yes, i have checked all possible permutations of the method, it really does work and i've compared it to the other subtraction methods and everything checks out.
I wonder if there's such an algorithm that does division in a reverse-multiplication way, or if there's an absurdly useless way to do addition using reverse-subtraction.
That would be so cool having a way to do a division in a format similar to addition, multiplication and subtraction.
I don't know if I'll ever find out if I am the true inventor of this, but if anyone has answers about this then let ke know :D.
gosh im so much slower at addition now that im not in school but this trick is legendary
Is there are similer strategy for multiplication/devision
took about 5 seconds to compute the thumbnail is 1861, though when i force myself to go fast i often make errors, like accidentally getting an answer 200 below what the actual answer is
I only looked at the thumbnail and got 2061 as the answer.
Yayy I got ittt 🎉🎉🎉🎉🎉🎉🎉
This was so easy and effective I did it to my IQ. 👍
I had this form of subtraction in my year 2 math workbook but my teachers let all of us use borrowing instead because it’s probably easier to use
It’s such a shame your teacher couldn’t be bothered to learn this more efficient method so just taught you what she already knew.
I didn't know that I use that but in most cases I do
I got 2059 in a few seconds from going right to left with 2100 and 58 - 97
ngl I read right to left (Arabic reader reading Arabic numerals).
It's obvious - add 3 to 2597 to get 7400, then add 2000 to get 9400, then add another 58. 3+2000+58 = 2061.
The key to getting good at arithmetic is getting comfortable with the value of numbers and then making life easy for yourself in calculating.
You could also get the answer easily by 9458 - 7400 is 2058 and then add 3 is 2061.
The hard way is to plod through a method by rote, that gives you the right number but, yes, doesn't ever ENGAGE with the values of those numbers.
As for 1385 + 3749, add 15 to the first, taje it away from the second and you get 1400 + 3734 which is 5134.
Oh and I HATE the way you write 4s.
I hate the way you type your 4s
I always work left to right. It pisses my wife off.
Last time I came this early, my girlfriend dumped me
HA, NERD
Me too
Hey man, can you help me demonstrate whether lagrange's theorem applies to n-ary groups or not?en.wikipedia.org/wiki/N-ary_group
and happy new years btw