Hi fellow Brilliantians out there! As any kid studying for a competitive exam, an arsenal of vast number of tricks, which make calculations simple, is a must. Why not try & write some of these here in this feed?

Here are some of mine -

Do it this way!

\[11 \times 23 = 2 (2+3) 3 = 253\]

Hey presto! You got the right answer!

Take another one- \[11 \times 49 = 4 (4+9) 9 = 4 (13) 9 = (4+1) 3 9 = 539\]

Just Brilliant! (Pun intended).

Take the case of \(145^2\).

**1.** Take the number other than 5 & consider it as a whole (in this case 14).

**2.** Multiply it by its consecutive integer (here \(14 \times 15 = 210\)).

**3.** Join 25 at the back of the number you just got (here 21025).

And that, my dear friend, is the answer! (Check if you don't believe me!).

**Remember**: This is only valid for the squares of integers ending with 5.

I hope to add more!

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## Comments

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TopNewest@Ameya Salankar I know the fastest trick to find the cube root of any number.(But you should know that the number is a cube.)

Eg- Find the cube root of 474552

Units digit is 2. This means that the units place of the cube root will be 8. (\(8^3 = 512.\))

Now, cancel the last 3 digits of the number. Hence, we are left with 474.(You have to always cancel the last 3 digits)

Now, you have to find the cube which is smaller than 474 but closest to 474. It is 343. 343 is \(7^3\).

Therefore, our required answer is \(\boxed{78}\).

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You forgot the 8 after the 7. It should be 78. Nice one though!

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Oh! Thanks

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Oh! I knew this one! I kinda forgot... (I am

knownfor my forgetfulness!)Log in to reply

just came across this image .Hope you may like it

abc

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déjà vu (its real this time!). By the way, which book is that? (The trick on the next page....?)

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I have actually taken it from my gmail account.somebody had posted it

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Wow! This is a nice trick.

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So how does this work?

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@Ameya Salankar Thanks for tagging me.

Now that you have given a way to square numbers ending with a \(5\), here is a general way to square numbers ending with any digit.

\({ a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b)\)

\({ a }^{ 2 }=(a+b)(a-b)+{ b }^{2}\)

So for example if you wanted to calculate the square of \(23\), then

\({ 23 }^{ 2 }=(23+3)(23-3)+{ 3 }^{ 2 }=20\times 26+9=\boxed{529}\).

I prefer this method because doing (\(26\times 20\)) is easier than doing (\(23\times 23\))

The method provided by Ameya to square numbers ending in \(5\) is also based on the above identity.

\({ 145 }^{ 2 }=(145+5)(145-5)+{ 5 }^{ 2 }=140\times 150+25=21000+25=\boxed{21025}\)

Hope this helped :)

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A treasure box of math tricks

Just browsing Arvind gupta toys.com site , a very good site containing interesting and valuable information on all topics. Just click this link

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This is not exactly ground-breaking, but it helps me.

To find \(a^2\), given \((a-1)^2\), just add \((a-1) and (a)\) to get the desired result. This can be repeated many times to get \(a^2\) from \((a-n)^2\)

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Another way to find square of any two digit number:- Consider a two digit number AB so we can write it as:- AB=10A+B now squaring both sides:- \[(AB)^2=100A^2+10AB+B^2\] Now this equation can be used to find square of any two digit number eg:- To find square of 16:-

1)find the square of last digit and write one's digit of answer obtained and take ten's digit as carry for next step(in this case square of 6 is 36 so write 6 in your answer and take 3 as carry)

2)multiply both digits of number with each other and 2 and add the carry obtained in previous step(in this case 1×6×2+3) Now write one's digit of number obtained in second step to left of number obtained in first step and take ten's digit of number obtained in second step as carry(in this case i obtained 6 from first step and 5 from second step so write ....56 also i got 1as carry from second step)

3)now find square of ten's digit and add the carry obtained in second step to it and write result obtained with your answer(in this case \(1^2+1=2\) and write it to left of your answer (256) ) I hope my wordings make sense

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Let me tag some of you here

@Krishna Ar,Sharky Kesa, Finn Hulse , Anik Mandal, milind prabhu , Calvin Lin , Avineil Jain , Aditya Raut , Satvik Golechha Agnishom Chattopadhyay ,Daniel Liu , Nanayaranaraknas Vahdam @Dinesh Chavan @Sreejato Bhattacharya

Oh no! This list is getting too long!

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That is nice. Why don't you create a wiki page called Vedic Math? Many people could put together their knowledge since wiki pages are collaborative.

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Yeah. That's right. You must consider it Ameya...(Back after a long break?..How was Kvpy?) and you too agnishom ...how was it?

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@Krishna Ar

Forget KVPY. I couldn't believe that was Ameya Salankar giving KVPY like that!Fell face down! I am going to try harder next year.

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@GAUTAM SHARMA, my marks are out of hell! You first....

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(Just Joking but the marks are real)

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@Krishna Ar Yes. And I want the same treatment (shocked & gaping) when I bring near perfect marks next year....(just being realistic!)

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@Krishna Ar, thanks! What about your NMTC?

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To square any number ending with the number six. For example 36

36 Step 1: Attach 2 to the number before six . (The number here is 3, so attaching 2 to 3, we get 32) Step 2: Multiply the result by 1+the number before six (here it corresponds to 1+3=4. so result times this = 324 =128) Step 3 : Add 1 to this new result (we get 128+1=129) Step 4: attach 6 to the answer. (we get 1296)therefore, 36*36=1296. I know its very complicated. It might be be useful for smaller numbers though

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Wo, well about 11, I'm JUST MAD!!!

Everytime what I'm doing is making words into numbers and finding their \(\pmod{11}\) value ! Almost all words I come across, I think of them as numbers and then try some arrangements in them like giving space so that it becomes divisible by 11 :P Sounds insane but that's my most favorite timepass :P

And my love about "11" is also the reason why this set was made ^_^

About tricks, I've got an awesome file, which has many many many tricks like the ones in your note... I hope you'll like them ... (On google, shared with all who have the link, here's the link - Maths PDF )

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If you want to find out square root of a number (approx.) then use this vedic math method:

sqrt(any number)=sqrt(nearest square) - {(nearest square)-(original number)}/[2 x sqrt(nearest square)]

For example: we want sqrt(22) then find nearest square of it that is 25. Now, sqrt(22)=5-(25-22)/(2 x 5)=5-3/10=4.7 Actual value of sqrt(22) is 4.69

NOTE:sqrt(a) means square root of 'a'

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Actually I didn't quite understand this trick. Could you please use \(\LaTeX\)? @Vaibhav Jain

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Given something symmetrical you have to multiply or add, pair opposite up.

Example: Evaluate \(\sqrt{12\cdot 13\cdot 14\cdot 15+1}+\sqrt{16\cdot 17\cdot 18\cdot 19+1}\)

Pair up opposites and multiply: \(\sqrt{(12\cdot 15)\cdot (13\cdot 14)+1}+\sqrt{(16\cdot 19)\cdot (17\cdot 18)+1}\)

Difference of squares: \(\sqrt{(180+1)^2-1^2+1}+\sqrt{(304+1)^2-1^2+1}\)

Simplify: \((180+1)+(304+1)=\boxed{486}\)

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Hey I am having PSA on this 20th please give me some tips or tricks to help me!

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For PSA (assuming you are in 9th) there is no need for special tricks but i have some tips for you- 1. Carefully read all paragraphs related to questions because most answers are in them specially in science and formulae are specified in text or you can derive them easily in maths.

2.There is no specified section for social studies but there is a section related to moral science and marks for S.St are calculated from this section. 3.English is quite simple and most of the reference can be taken from paragraphs. 4. No need to hurry because their is plenty of time .

BEST OF LUCK> thats all.

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A way to find that a number(not very large) is prime or not is that take square root of that number and check its divisibility by prime no. less then square root, if it is not divisible by primes less than square root the given no. is prime.

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If anybody wants to learn some divisibility rules he can do it here

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An easy method for finding the digital root of any number is to cast out \(9's\) and the group's of digits which add up to nine.This is done by crossing out any nines in the number or any two digits adding up to \(9\).The numbers which are left at the end give the digital root of the number.If there is nothing left after casting out the nines,then the digital root is\(9\).

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Wow. Awesome I am trying it

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the sum of positive odd numbers is given by n>2 ; 1 +3 =4 ;2

2 1+3 +5 =33 1+3 +5 +7 +9 =25 ; 5*5 just count odd numbers. square it.you will get sum of odd numbers.Log in to reply

Well I have another method to find the product when a number is multiplied by 11.Here it goes:

Suppose you have to multiply \(13,423\)by \(11\)

Step 1:Write down the number with a nought placed on both ends.

Step 2:Add the final two digits to get the units digits of the product.

Step3:For the tens digit,add the final two digits at that point and continue in this fashion to get the product.

For example,\(013,4230\)-Units digit \(0+3=3\)

Continue this way to get the product as \(147,653\)

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lets see....To find the square of any number ending with 5. like 25x25=(2x3)25 i.e.625 65x65=(6x7)25i.e.4225 so, basically n5xn5=(n x (n+1))25 Try it out

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State and prove thales theorem?

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