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Easy Method of Calculating The Square Root of Huge PERKS

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sirjon Just gettin' started! Default Avatar
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Easy Method of Calculating The Square Root of Huge PERKS Post Tue Jul 20, 2010 10:48 pm
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    I am searching in Google about how to calculate the square root of numbers in an easier way beside the long hand division method. I found one, which is an easy instruction by Z-math. But I found it, seems tricky when the perks (a short name for ‘perfect square numbers’), involved, are in 6 or in 8 digits. I developed a much easier method, I called MSM-1 and would be glad to share to you. First, I recommend you to read Z-math instructions http://www.ehow.com/how_2322332_square-root-number-mentally.html (if the Admin permit it), to shorten the discussion.

    Let’s start with a 6-digit perk, example \/66,564.

    Step 1: Regroup the digits by twos

    \/06’65’64

    Step 2: Find the nearest square, equal or less than the first group of digits (that is, 06). Write down below it the equivalent square root.

    \/06’65’64
    2

    Step 3: Find two squares that ends with 4. There are always a pair, 4 and 64.
    Write down their equivalent square roots, 2 and 8.


    \/06’65’64
    2 _ 2
    2 _ 8

    Step 4: Notice that our problem is, what would be the middle digit? There are still ten choices, 0, 1, 2, 3, 4, 5, 6, 7,
    8 and 9.

    Step 5: Write down N as the missing digit.


    \/06’65’64
    2 N 2
    2 N 8

    Step 5: Apply the SSQ Method.


    SSQ METHOD

    SSQ stands for ‘Systematic Squaring’. It is based on this popular algebraic expression, (x + y)^2 = x^2 + 2xy + y^2.

    But to simplify the presentation, I recommend that all ‘index squares’ should be in two digits, so instead of 1^2 = 1, it should be 1^2 = 01. The same as to 2^2 and 3^2, they should be 04 and 09, respectfully. There are three main parts for SSQ, namely - PSL (Partial Square Line), SP1, SP2, SP3 and so on (Sub-product or sub-products that depend on how many digits involved, in squaring that number), the ST1, ST2, SE3 and so on (sub-totals, again, depend on how many digits involved) and TSUM (or the total sum).
    ...TO BE CONTINUED

    (Note: I cannot find the square and sqaure root signs , so simply consider \/ as square root sign and ^2 as to the power of 2 or 'square of'.)

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    Last edited by sirjon on Sat Jul 31, 2010 2:46 am; edited 2 times in total

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    sirjon Just gettin' started! Default Avatar
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    Post Wed Jul 21, 2010 7:35 pm
    I will try my very best, to explain to you things in the easiest, possible ways I could. Maybe you’re asking yourselves, what is the significance of SSQ? SSQ is the key to easily understand, SE or Square Edging, a counter part method of SSQ that I devised to find the square roots of large valued perfect squares in a much easier way.

    Let us consider squaring the number 123

    123^2

    Step 1: Create a PSL (Partial Squares Line)

    The PSL is simply, the “two-digit squares” representation of each, individual digits of a certain number. In 123^2, the two-digit squares representation of 1, 2 and 3 are 01, 04 and 09, respectively.

    123^2 = 01’04’09

    Algebraically, 123^2 can be represented by this equation, (W + X + Y)^2, provided that W = 100, X = 20 and Y = 3.

    (W + X + Y)^2 = {W + (X + Y)}^2 = W^2 + 2W(X + Y) + (X + Y)^2

    = W^2 + 2W(X + Y) + X^2 + 2XY + Y^2

    The PSL value, 01’04’09 or 10,409 to be exact, is equal to the sum of W^2 + X^2 + Y^2

    It leaves 2XY and 2W(X + Y).

    Step 2: Solve SP1 and SP2

    SP stands for Sub-product. In a 3D.SSQ (three-digit Systematic Squaring), there always appears two sub-products.

    SP1 = 2XY

    SP2 = 2W(X + Y)

    Solving SP1 and SP2 algebraically,

    SP1 = 2 x 20 x 3 = 120

    SP2 = 2 x 100 x (20 + 3) = 2 x 100 x 23 = 4600 or 4,600

    But in SSQ, there is a short cut method to solve SP1 and SP2

    General Rules in Dealing with Sub-products

    Rule 1: Cross multiply the individual digits of a given number using the R.A.R. multiplication pattern

    Rule 2: Don’t forget the DTP reminder, “Double The Product”

    Rule 3: Follow the decimal place of the reference digit


    R.A.R. Multiplication Pattern

    R.A.R. stands for “Reference Digit and All the Digits to the Right”. It is a multiplication pattern that is effective in solving the sub-products. How it works?

    Consider 123^2

    First Pattern:

    If we pick 3 as our reference digit, ‘the all to the right’ of 3 is nothing, null, empty or zero. Multiplying 3 by 0 is futile, so, we can skip this pattern.

    Second Pattern (SP1):

    If we pick 2 as our reference digit, ‘the all to the right’ of 2 is 3.

    Rule 1: 2 x 3 = 6

    Rule 2: DTP reminder 6 x 2 = 12
    (Note: rule 2 is optional, you can skip it, as long as you would no to forget to include the “x 2” in rule 1, that is, 2 x 3 x 2 = 12

    Rule 3: Our reference digit 2 is in the tens decimal place, therefore, the last digit of SP1 must be also, in the tens decimal place.

    SP1 = 12 (provided that 2 is aligned to the tens decimal place)

    Third Pattern (SP2):

    If we pick 1 as our reference digit, ‘the all to the right’ of 1 are 2 and 3. But we must consider the digits 2 and 3 as a group and not as individual digits, the very reason why I gave emphasis to the word ‘ALL’ in that phrase, “All to the Right”.

    Rule 1: 1 x 23 = 23 Correct
    1 x 2 x 3 = 6 Wrong!

    Rule 2: DTP reminder 23 x 2 = 46

    Rule 3: Our reference digit 1 is in the hundreds decimal place, therefore, the last digit of SP2 must be also, in the hundreds decimal place.

    SP2 = 46 (provided that 6 is aligned to the hundreds decimal place)

    Sub-Totals

    A ‘sub-total’ is simply, a temporary sum, in between, each time you add a sub-product. In squaring two-digit number (such as, 142), there is no sub-total because you directly add SP1 to the PSL. Sub-total is optional, you can either include or ignore it. But personally, I highly recommend you to practice including it because it is important in studying (or analyzing) Square Edging.

    In 3D.SSQ, there is only one sub-total. It is the temporary sum PSL + SP1

    123^2 = 01’04’09 ← PSL
    + 2x3x2 = 12 ← SP1 (provided that 2 is aligned to the third zero of the PSL)
    01’05’29 ← ST1 (provided that the 9 is aligned to the last digit of the PSL)

    Total Sum

    The total sum is the final answer. It reflects the ‘true’ square value of a given number. If you multiply 123 by 123, using the common method of multiplying numbers, you will discover that the total sum of SSQ method is exactly equivalent to the product of 123 x 123.

    T-Sum = ST1 + SP2

    01’05’29 ← ST1 (provided that the 9 aligned to the last digit of the PSL)
    +1x23x2 = 46 ← SP2 (provided that 6 aligned to 5 of ST1)
    01’51’29 ← T-SUM (provided that 9 aligned to the last digit of ST1)

    SUMMARY:

    123^2 = 01’04’09 ← PSL
    + 2x3x2 = 12 ← SP1 (provided that 2 aligned to the third zero of the PSL)
    01’05’29 ← ST1 (provided that the 9 aligned to the last digit of the PSL)
    +1x23x2 = 46 ← SP2 (provided that 6 aligned to 5 of ST1)
    01’51’29 ← T-SUM (provided that 9 aligned to the last digit of ST1)

    Important Note: Due to HTML problem of aligning the digits in their proper decimal places (I tried many times but failed to resolve it), I included notes on the right side of each mathematical sentences.

    More to come...soon.














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    Last edited by sirjon on Sat Jul 31, 2010 2:43 am; edited 1 time in total

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    sirjon Just gettin' started! Default Avatar
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    Post Sat Jul 31, 2010 2:39 am
    Ok let me continue Square Edging \/66,564.

    \/6'65'64.
    2 N 2
    2 N 8

    square of 2N2 = 04'nn'04 PSL
    Last digit of SP1 = ..6 -..0 = ..6 (digit at tens dec. place of TSum minus the digit at tens dec. place of PSL)
    Nx2x2 = ..6
    Nx4 = ..6 N = 4, 9

    Possible Sq. Rts = 242, 292 (digit at tens dec. place of TSum minus the digit at tens dec. place of PSL)

    square of 2N8 = 04'nn'64 PSL
    Last digit of SP1 = ..6 -..6 = ..0
    Nx8x2 = ..0
    Nx16 = Nx6 = ..0 N = 0, 5

    Possible Sq, Rts. = 208, 258

    We have four possible square roots and only one is the true sq. rt.

    Using a Parameter Checker, I detected that the 6'65'64 is in the Higher Limit Area

    Two Possible Sq. Rts = 292 and 258

    Using a Square Root Locator,

    H = 9'00'...
    M = 7'56'..
    L = 6'25'..

    6'65.. is in-between 6'25'..and 7'56'..therefore the true square root is in the lower side, therefore the true square root is 258

    If you can't understand how I did it, I posted a blog about it but I will ask first the permission from the admin.

    The true challenge of S.E. (MSM-1 format) is when square edging 8-digit perfect square numbers.

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    Abhishek009 Really wants to Beat The GMAT! Default Avatar
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    Post Tue Aug 03, 2010 4:26 am
    sirjon wrote:
    I am searching in Google about how to calculate the square root of numbers in an easier way beside the long hand division method. I found one, which is an easy instruction by Z-math. But I found it, seems tricky when the perks (a short name for ‘perfect square numbers’), involved, are in 6 or in 8 digits. I developed a much easier method, I called MSM-1 and would be glad to share to you. First, I recommend you to read Z-math instructions http://www.ehow.com/how_2322332_square-root-number-mentally.html (if the Admin permit it), to shorten the discussion.

    Let’s start with a 6-digit perk, example \/66,564.

    Step 1: Regroup the digits by twos

    \/06’65’64

    Step 2: Find the nearest square, equal or less than the first group of digits (that is, 06). Write down below it the equivalent square root.

    \/06’65’64
    2

    Step 3: Find two squares that ends with 4. There are always a pair, 4 and 64.
    Write down their equivalent square roots, 2 and 8.


    \/06’65’64
    2 _ 2
    2 _ 8

    Step 4: Notice that our problem is, what would be the middle digit? There are still ten choices, 0, 1, 2, 3, 4, 5, 6, 7,
    8 and 9.

    Step 5: Write down N as the missing digit.


    \/06’65’64
    2 N 2
    2 N 8

    Step 5: Apply the SSQ Method.


    SSQ METHOD

    SSQ stands for ‘Systematic Squaring’. It is based on this popular algebraic expression, (x + y)^2 = x^2 + 2xy + y^2.

    But to simplify the presentation, I recommend that all ‘index squares’ should be in two digits, so instead of 1^2 = 1, it should be 1^2 = 01. The same as to 2^2 and 3^2, they should be 04 and 09, respectfully. There are three main parts for SSQ, namely - PSL (Partial Square Line), SP1, SP2, SP3 and so on (Sub-product or sub-products that depend on how many digits involved, in squaring that number), the ST1, ST2, SE3 and so on (sub-totals, again, depend on how many digits involved) and TSUM (or the total sum).
    ...TO BE CONTINUED

    (Note: I cannot find the square and sqaure root signs , so simply consider \/ as square root sign and ^2 as to the power of 2 or 'square of'.)
    Here are a few more links of quicker maths :


    SQUARING

    http://www.youtube.com/watch?v=sjgn8S_1i_g&playnext=1&videos=n7wwHCw3mYU&feature=rec-LGOUT-exp_fresh%2Bdiv-1r-9-HM

    http://www.youtube.com/watch?v=RzSHEDdoUco&NR=1

    http://www.youtube.com/watch?v=I9t-gYnPNaw&feature=related






    CUBE ROOTS

    http://www.youtube.com/watch?v=gIo88Rt1JtI&feature=related

    http://www.youtube.com/watch?v=zlM_OKJtnXM&feature=related





    FINDING CUBE ROOTS AND SQUARE ROOTS OF IMPERFECT SQUARES AND CUBES

    http://www.quickermaths.com/herons-method-of-finding-roots/

    _________________
    Abhishek

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    sirjon Just gettin' started! Default Avatar
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    Post Mon Aug 09, 2010 2:28 am
    Abhishek009 wrote:
    Thank you for the reply. Actually, I'm not asking for it but rather wish to share an easy method of getting the square roots of numbers, even in eight digits in an easier way (of course, provided that they are in perfect squares). I'm working on it but will ask first the admin if I can post it here. Anyway, thanks a lot

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    sirjon Just gettin' started! Default Avatar
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    Post Fri Aug 20, 2010 1:19 am
    I already finished my blog Square-Edging (MSM-1 Format). I wish to share my blog, "Easy Square Root method for Grade school Kids" but I don't know if I might violate both the rules of GMAT and Google's blogspot. I'm asking the admin if it is possible to do it here? Any suggestions or help? Thanks a lot.

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    beastly B Just gettin' started!
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    Post Mon Dec 27, 2010 8:55 pm
    sorry i do not understand its a bit complicated Question

    sirjon Just gettin' started! Default Avatar
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    Post Tue Dec 20, 2011 6:03 pm
    Modifying SSQ

    I realized that it is 'important' how to know squaring a number that in the process, would be advantageous if one wants to learn MSM-3 (Easiest Method of Taking the Square Root of any number, large or small. perfect or non-perfect squared numbers)

    Let's take our example, squaring 123.

    Based on this popular equation...

    (X+Y)^2 = X^2+2XY+Y^2

    We could first deal with the first two digits of 123, that is 12...

    12..^2 = 01'04.. (Partial Square Value, equivalent to X^2+Y^2, provided that X=10 and Y=2)
    1x(2x2)=+__4_ (Sub-product which is equivalent to 2XY)
    TSquare=01'44.. (True Square Value)

    Therefore 12..^2= 144..

    The two dots ( .. ), signify that the value is still incomplete, provided that 1 and 2 are just the first two digits of 123, therefore 144.. only signify a temporary value.

    123^2 can be based on this algebraic equation...

    (X+Y+Z)^2= {(X+Y)+Z}^2 = (X+Y)^2+2Z(X+Y)+Z^2

    We already solved (X+Y)^2 = 1'44.. (provided X=10 and Y=2)

    Therefore,

    123^2 = 1'44'09 (PSV, provided X=100, Y=20 and Z=3)
    12(3x2)=___7'2_ (SP)
    TSquare=1'51'29

    Now if there is added digit or digits such as...

    1238^2 = ?

    Use the basic algebraic equation for (X+Y)^2 = X^2+2XY+Y^2

    Provided that X = 123.. and Y = 8.

    Solved it for yourself and try if your answer is correct using a 'calculator'.

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