@MegaByte • 18 May, 2011

We all being Engineers have the habit of solving most of the problems using 'Calculator'. So, let us share the tricks to perform the basic calculations without making the use of calculator.

I make the start.

So the final answer turns out to be

If this approach is practised for 2-3 times, then surely multiplication of 2-digit numbers becomes quicker than that of the traditional method. Just give it a try. 😉

I make the start.

**To multiply any two, 2-digit numbers**:- Multiply the unit-place digits of both the numbers and note it down as a part of result.
- Cross multiply the digits and add their product.
- Multiply the tens-place of both the numbers.

**Eg:**To multiply 41 and 62:**Step 1:**Multiply (1*2) i.e units place of both numbers which comes out to be

**2**.

**Step 2:**Cross multiply both numbers. Multiply (4*2) and (1*6). Add their products i.e. (8+6=1

**4**). Here, 1 is considered as carry.

**Step 3:**Multiply (4*6) i.e. tens place of both numbers which comes out to be 24. Now add the carry (1) to 24, which gives out

**25**.

So the final answer turns out to be

**2542**. You can check it on calculator. 😛

@CIVILPRINCESS • 18 May, 2011
i learnt this method of multiplication in my training class! its very useful 😀 even for a 4digit by 4 digit multiplication it is useful 😁

@Ankita Katdare • 18 May, 2011
Awesome thread MegaByte! 😀 I think we should come up with more tricks that come handy.

How about starting some threads where we have tutorials for simple vedic mathematics?

How about starting some threads where we have tutorials for simple vedic mathematics?

@ISHAN TOPRE • 18 May, 2011
Very good megabyte. I have some tricks too. For adding numbers which has a same MSG and the sum of LSB is 10.

For example 43X47 . Here The Most significant digits are same i.e.; 4 and addition of ten's place i.e.; least significant digits is 3+7=10.

So, 7X3=21

The add 1 to one of the 4. to make it 5. Now do 5X4=20.

So we have 43X47=20 21

Simple isn't it?

Now try for 62*68, 85*85, 96*94. 😁

For example 43X47 . Here The Most significant digits are same i.e.; 4 and addition of ten's place i.e.; least significant digits is 3+7=10.

So, 7X3=21

The add 1 to one of the 4. to make it 5. Now do 5X4=20.

So we have 43X47=20 21

Simple isn't it?

Now try for 62*68, 85*85, 96*94. 😁

@MegaByte • 18 May, 2011

Yes, the same method can used to for 3-digit and 4-digit multiplication.CIVILPRINCESSi learnt this method of multiplication in my training class! its very useful 😀 even for a 4digit by 4 digit multiplication it is useful

@MegaByte • 19 May, 2011

Thats a nice idea. We can surely come up with the basics of Vedic Mathematics. 😀AbraKaDabraHow about starting some threads where we have tutorials for simple vedic mathematics?

@MegaByte • 19 May, 2011

In order to add all the numbers in such series, it generally becomes very tedious to do it one-by-one. So, try out this method: (n * ( n + 1 ) ) / 2

Trick to add a series of naturals numbers upto 'n' :Trick to add a series of naturals numbers upto 'n' :

**1 + 2 + 3 + 4 +........ + 83**

Eg:Eg:

In order to add all the numbers in such series, it generally becomes very tedious to do it one-by-one. So, try out this method: (n * ( n + 1 ) ) / 2

= (83 * 84) / 2

= 3486

Now, if you wish to add all these numbers using calculator, try it out. You'll get the same result with more consumption of time. 😁
= 3486

@MegaByte • 19 May, 2011
@Ishu- Though the trick you've mentioned is number specific, it is very quick and easy. 😉

@MegaByte • 25 May, 2011

**:**

Multiplying any number by 5, 25, 125Multiplying any number by 5, 25, 125

Simply append 0, 00, 000 to the multiplicand (for 5, 25, 125 respectively) and then divide it by 2, 4, 8 respectively.

( 9897952 * 5 ) = ( 9897952

( 9897952 * 25 ) = ( 9897952

( 9897952 * 125 ) = ( 9897952

**Eg:**( 9897952 * 5 ) = ( 9897952

**0**/**2**) = 49489760( 9897952 * 25 ) = ( 9897952

**00**/**4**) = 247448800( 9897952 * 125 ) = ( 9897952

**000**/**8**) = 1237244000

@CIVILPRINCESS • 27 May, 2011

sum of squares upto n (1+4+9+16+...+n2) = (n*(n+1)*(2n+1))/6

sum of cubes upto n(1+8+27+....+n3)=((n*(n+1))/2)^2

for sum of natural numbers upto n = (n*(n+1))/2MegaByte

Trick to add a series of naturals numbers upto 'n' :

1 + 2 + 3 + 4 +........ + 83

Eg:

In order to add all the numbers in such series, it generally becomes very tedious to do it one-by-one. So, try out this method: (n * ( n + 1 ) ) / 2= (83 * 84) / 2Now, if you wish to add all these numbers using calculator, try it out. You'll get the same result with more consumption of time. 😁

= 3486

sum of squares upto n (1+4+9+16+...+n2) = (n*(n+1)*(2n+1))/6

sum of cubes upto n(1+8+27+....+n3)=((n*(n+1))/2)^2

@Saandeep Sreerambatla • 03 Jun, 2011
I would suggest every exam taker (like CAT and GMAT) donot remember most of the tricks, its waste of time.

Only practice those tricks which are easy and helpful. Since if you go to any coaching center , they will teach you many tricks but its not highly recommended.

Only practice those tricks which are easy and helpful. Since if you go to any coaching center , they will teach you many tricks but its not highly recommended.

@busoglusog • 17 Jun, 2011
very nice tecnique!!!

@Karshil Sheth • 26 Jan, 2020 • 1 like

I have difference of opinion here. Firstly if we are encouraging technology then why can't we use it often then calculating by hand. Also now a days this trick works in exams like school or mostly in jee where its difficult to calculate. In gate also they have provided virtual calculator, so I hope people will use it for its ease.

But the trick was nice to learn and I found it quite interesting.!

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