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Quick way to calculate square root of large number?

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calande Rising GMAT Star Default Avatar
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Quick way to calculate square root of large number? Post Tue Jun 03, 2008 5:54 am
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    Hi guys,

    Do you know how to calculate quickly √576 ?
    What's your method?
    Thanks Wink

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    gogetter08 Just gettin' started! Default Avatar
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    Post Tue Jun 03, 2008 7:13 am
    Unfortunately, I know from the top of my mind what the answer is since I am an engr with strong math skills.

    There is no direct shortcut for squareroot but here are 2 ways that you can try based on your skills: I suggest to go by elimination of choices

    method1:
    1. You can approximately get a range for the answer by seeing that 576 lies between 400 and 900(whose sq roots are 20 and 30).
    2. Now that we know that the answer is between 20 and 30, try to see what number should be in the units place to get a 6(576) in the product with itself. Either a 4 or 6 wd do that. So possible answers are 24 or 26.
    3. At this time you verify one by multiplying 24x24 to verify

    method2:
    1. Write down the factors of 576 and and see if they share the same with the choices. For eg, if you see 25 and 26, you can tell that there is no factor of 5 or 13 in 576 and eliminate them!

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    calande Rising GMAT Star Default Avatar
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    Post Tue Jun 03, 2008 8:09 am
    Thanks, it helped!

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    Post Tue Jun 03, 2008 10:41 am
    It's useful in many questions to know the small powers of the smallest primes. If you know that 5^4 = 25^2 = 625, you'll know that the positive square root of 576 is just less than 25, which leads you to the correct answer- 24- quite quickly.

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    Post Thu Jun 05, 2008 1:33 am
    Not that this is what your asking, but
    I think the non-short cut way of doing prime factorization is pretty fast in this particular case, and in most that I've seen.
    576/2 = 288
    288/2 = 144
    144= 12*12
    => 576=24^2

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    calande Rising GMAT Star Default Avatar
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    Post Thu Jun 05, 2008 1:52 am
    Thanks Jason. Your method works only for even numbers, right? I tried to calculate √529 with your method and dividing 529 by 2 doesn't yield an integer...How would you do in this case?

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    Post Thu Jun 05, 2008 2:41 am
    For finding the square roots of a perfect square number, it is advisable to remember the square of all integers from 1 to 30.
    The reason why I'm saying is that it works pretty faster if u need to find the square root of a non perfect square.

    Let's say u want to calculate how much is square root of 200.

    we know that square root of 200 would lie between 14(196) and 15(225).
    Now 200=14 + (200-196)/(225-196)
    =14+6/29=14.2(approx)

    As the number increases the probability of findind the exact square root increases. Try for smaller non perfect no.

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    Post Thu Jun 05, 2008 4:14 am
    dextar wrote:
    we know that square root of 200 would lie between 14(196) and 15(225).
    Now 200=14 + (200-196)/(225-196)
    =14+6/29

    dextar, that is not the *exact* square root of 200, it's only an estimate. The square root of any positive integer is either an integer or an *irrational number*, i.e. a number that cannot be written as a fraction involving two integers. root(200) = 10*(root(2)); put that in your calculator and you'll see it's different from 14+6/29.

    I like Jason's method above, but it will only be quick when the square root in question has a small prime factor (2, 3, or 5). If there is no small prime factor, I'd estimate, then use the last digit:

    root(400) < root(529) < root(625)
    20 < root(529) < 25

    Note now that 21^2 would end in 1, 22^2 would end in 4, 23^2 would end in 9, and 24^2 would end in 6. The only candidate among whole numbers is 23. Of course, there's no guarantee that the root is a whole number at all, but if you know you're looking for a whole number, this gets you there quickly.

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    Post Fri Jun 06, 2008 12:07 am
    Ian

    I agree with u that 14.2 is not the square root of 200. It's actually 14.414 . But as u increase the number the approximation aproaches the exactness!!!.

    If u really depends on these methods (For e.g suare root of 529 or say 841)for calculating even the square roots , I would say it is not advisable. The best way to remember these is learn squares of number up to 30. Life will be very smooth as far as calculation goes. Trust me

    The method I propsed will give u approx only.It will come handy if u really want to calculate some big numbers (not that big!!!) which are not the perfect squares.

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    Post Thu Jun 25, 2009 9:31 am
    how do you calculate the square root of a HUGE number, like 48400?
    i get to 20,/13, and then i'm stuck. help?

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    Post Thu Jun 25, 2009 9:42 am
    I own the GMAT wrote:
    how do you calculate the square root of a HUGE number, like 48400?
    i get to 20,/13, and then i'm stuck. help?
    If possible, divide it up into smaller numbers, the roots of which can be calculated easily. According to the square root rules:

    root(48400) = root(484) * root(100) = root(4) * root(121) * root(100)
    = 2*11*10
    = 220

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    Post Thu Jun 25, 2009 10:28 pm
    thers is also a simple method using prime factorization:
    the prime factors fo 576 are: 2^6 x 3^2

    so √576 is nothing but half of the powers of the prime factors i.e for every 2 consider 1
    √576 = 2^3 x 3 = 24

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    Post Wed Jul 15, 2009 3:39 pm
    I don't know if this helps but I found out another way to calculate the square of any two digit number.

    Ex 1:
    find 16^2
    =10^2 + 6^2 + (60*2)
    =100 + 36 + 120
    = 256

    Ex 2:
    find 19^2
    = 10^2 + 9^2 + (90*2)
    = 100 + 81 + 180
    = 361

    I was trying to find out a way to take squares of smaller digits inside of the larger number. Adding a 0 to the last digit and multiplying by 2 seems weird but this method works for every 2 digit number. Does anybody know how this method can be used for numbers greater than 2 digits? It seems like it could be useful for finding square roots and squares.

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    Post Wed Jul 15, 2009 5:06 pm
    fstwrx wrote:
    I was trying to find out a way to take squares of smaller digits inside of the larger number. Adding a 0 to the last digit and multiplying by 2 seems weird but this method works for every 2 digit number. Does anybody know how this method can be used for numbers greater than 2 digits? It seems like it could be useful for finding square roots and squares.
    You're actually just using this well-known factoring pattern (which shows up in all kinds of places on the GMAT) -

    (x + y)^2 = x^2 + 2xy + y^2

    So, using this with x = 10, y = 9, we have:

    19^2 = (10 + 9)^2 = 10^2 + 2*10*9 + 9^2 = 10^2 + 9^2 + 2*90

    Of course you can use that with any number at all; it's just that the calculations get messier the more complicated the number. Still, something like 101^2 is easy enough to compute:

    101^2 = (100 + 1)^2 = 100^2 + 2*100*1 + 1^2 = 10,201

    It's still going to be awkward to calculate something like 387^2, but you don't ever need to do that type of thing on the GMAT anyway, fortunately (it would be quite a boring test if you did!).

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    Post Tue Apr 27, 2010 10:26 am
    The square root of a number is just the number which when multiplied by itself gives the first number. So 2 is the square root of 4 because 2 * 2 = 4.

    Start with the number you want to find the square root of. Let's use 12. There are three steps:


    1. Guess
    2. Divide
    3. Average.

    ... and then just keep repeating steps 2 and 3.

    First, start by guessing a square root value. It helps if your guess is a good one but it will work even if it is a terrible guess. We will guess that 2 is the square root of 12.

    In step two, we divide 12 by our guess of 2 and we get 6.

    In step three, we average 6 and 2: (6+2)/2 = 4

    Now we repeat step two with the new guess of 4. So 12/4 = 3

    Now average 4 and 3: (4+3)/2 = 3.5

    Repeat step two: 12/3.5 = 3.43

    Average: (3.5 + 3.43)/2 = 3.465

    We could keep going forever, getting a better and better approximation but let's stop here to see how we are doing.

    3.465 * 3.465 = 12.006225

    That is quite close to 12, so we are doing pretty well.

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