Wanna Share Some Math Tricks?

@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 $\boxed{78}$.

just came across this image .Hope you may like it

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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 :)

A treasure box of math tricks

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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!

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

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 $(a-n)^2$

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

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'

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 )

To square any number ending with the number six. For example 3636 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

the sum of positive odd numbers is given by n>2 ; 1 +3 =4 ;22 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.

Wow. Awesome I am trying it

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$9$.

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

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.

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

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 $0+3=3$

                                                        Tens digit=\\\(3\+2=5\\\)

                                                         Hundreds digit =\\\(2\+4=6\\\)

Continue this way to get the product as $147,653$

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

State and prove thales theorem?