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  • Solving Mathematics Without Calculator

    MegaByte

    Member

    Updated: Oct 27, 2024
    Views: 1.8K
    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.

    To multiply any two, 2-digit numbers:

    1. Multiply the unit-place digits of both the numbers and note it down as a part of result.
    2. Cross multiply the digits and add their product.
    3. 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=14). 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. 😛
    ​
    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. 😉
    ​
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  • CIVILPRINCESS

    Member • May 18, 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 😁
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  • Ankita Katdare

    Administrator • May 18, 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?
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  • ISHAN TOPRE

    Member • May 18, 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. 😁
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  • MegaByte

    Member • May 18, 2011

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

    Member • May 18, 2011

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

    Member • May 18, 2011


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

    Eg:     1 + 2 + 3 + 4 +........ + 83

    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. 😁
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  • MegaByte

    Member • May 18, 2011

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

    Member • May 25, 2011


    Multiplying 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.

    Eg:
    ( 9897952 * 5 ) = ( 98979520 / 2 ) = 49489760

    ( 9897952 * 25 ) = ( 989795200 / 4 ) = 247448800

    ( 9897952 * 125 ) = ( 9897952000 / 8 ) = 1237244000
    ​
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  • CIVILPRINCESS

    Member • May 26, 2011

    MegaByte

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

    Eg:     1 + 2 + 3 + 4 +........ + 83

    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. 😁
    for sum of natural numbers upto n = (n*(n+1))/2
    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
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  • Saandeep Sreerambatla

    Member • Jun 3, 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.
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  • busoglusog

    Member • Jun 17, 2011

    very nice tecnique!!!
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