taking a square root means simply finding pairs of factors and, when you find a pair, moving one of the pair out of the root sign and eliminating the other SQRT4 = 2 because 4=(2)(2) - we find the pair of factors and move the 2 out of the root sign.
For bigger numbers this can be achieved by finding all the prime factors. I use a factor tree - to find that 576 has 2,2,2,2,2,2,3,3 as factors. If we isoate the pairs and take them out of the square root you get (2)(2)(2)(3) as the square root - or 24 as your answer.
if the number is 1152 - then you have 2,2,2,2,2,2,2,3,3 - there is one 2 that does not have a pair so the answer would be 24SQRT2 - if you have to solve to a decimal (unlikely on the GMAT) you can memorize your smaller square roots to approximate the answer.
Quick way to calculate square root of large number?
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sirjon
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I devised three methods of finding the square roots of numbers, large or smallI 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?
MSM-1, Square Edging "Pairing" Method - taking the square roots of large numbers up to 8 digits (that's in million) but it is limited to perfect square numbers only and not ending with 25 (Example, 13,498,276 does not end with 25)
MSM-2, Square Edging "Grouping" Method - to complete the square edging method for large number perfect squares, ending with 25 (Example, 21,390,625). I realized that MSM-1 does not applied to 'perks' (perfect square numbers), ending with 25.
MSM-3 Universal Square Edging Method (U.SE) - using an innovative technique which I called NT-SSQ (Non-Terminating Systematic Squaring) and math tools like the P-CHk (Parameter Checker), D-Loc (Digit Locator) and T-Sampling (Test Sampling), I can calculate any number, even non-perfect squares without the use of a calculator and what so called 'estimation' method. This method is direct and systematic, without any trial and error method, very little division (division of 2 only) and no large number subtraction (single numbers subtraction only).
I will share the MSM-1 method but I rather reserve MSM-2 and MSM-3, for some security reasons (intellectual property). I'm intended to post MSM-1 later (or sooner) as a new thread. Thanks
- fatalityish
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The method discussed by the Swamiz is the age old Babylon Method of calculation of square root. It is effective but also time consuming. I feel the long division method is the best way to do this. It also helps in case we need to find the square roots of numbers with decimals.
With a little practice it can be expertized.
I dont think GMAC would ever give a sum so direct as to asking for the square root of a perfect square and in babylon method we still have some division to make like for example while calculating square root of 12, Swamiz divided 12 by 3.5. Further ahead you go it gets more complicated.
I got the following link while surfing the net which gives the square root algorithm explanation by guess and check method and long division in easy steps:-
https://www.homeschoolmath.net/teaching/ ... orithm.php
The best thing is this procedure can be applied for cube roots/4th roots/5th roots.... and so on.
With a little practice it can be expertized.
I dont think GMAC would ever give a sum so direct as to asking for the square root of a perfect square and in babylon method we still have some division to make like for example while calculating square root of 12, Swamiz divided 12 by 3.5. Further ahead you go it gets more complicated.
I got the following link while surfing the net which gives the square root algorithm explanation by guess and check method and long division in easy steps:-
https://www.homeschoolmath.net/teaching/ ... orithm.php
The best thing is this procedure can be applied for cube roots/4th roots/5th roots.... and so on.
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sirjon
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This is simply to update members of this forum: Square Root: A Digit-Per-Digit Approach https://sqrtdigperdig.blocked/
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sirjon
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beastly B wrote:what may be the esiest way to find the square root thats not complicated
Thank you Beastly B for the comment.
I hope this time, taking the square root of number can be taught in elementary schools without giving these poor children too much hard time.
I'm sure they'll enjoy it much better and they may develop their skills in adding and multiplying numbers much better and faster, learning four things in one package (adding, multiplying, squaring and square rooting).
Simply call this technique, MSM3, as a token of appreciation...short for Mendoza's square root method, version 3.
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chriscurtisrzf
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I'm currently working with numbers of arbitrary length (2048bit+).
I've written some C# code for dividing numbers of arbitrary length. Numbers as huge as the biggest RSA numbers take less than a second to divide by any number on a normal laptop.
Lemme know if you'd like a copy of the code. Always happy to share the shortcuts!
Background of current work...
Having established a pattern in the binary digits of prime pairs and the fractal relation to each set of prime pairs from the first pair (three and five) and their relation to bin:01 and bin10 (binary:one the 'zeroth' prime and two, the only even prime), I'm hoping the rest of the stuff I'm doing is going to be able to reduce the intractible factorisation of large prime products and integers to a (sub)polynomial time or trivial exercise on a normal PC in O(N). Working from the big end of any binary number, the carries of the previous smaller bits are already determined. This is VERY helpful... a solution to this factorisation intractibility is a BIG indicator that (counter-intuitively) P = NP and suggests we don't have our numbers organised in a 'logical' way...
I've written some C# code for dividing numbers of arbitrary length. Numbers as huge as the biggest RSA numbers take less than a second to divide by any number on a normal laptop.
Lemme know if you'd like a copy of the code. Always happy to share the shortcuts!
Background of current work...
Having established a pattern in the binary digits of prime pairs and the fractal relation to each set of prime pairs from the first pair (three and five) and their relation to bin:01 and bin10 (binary:one the 'zeroth' prime and two, the only even prime), I'm hoping the rest of the stuff I'm doing is going to be able to reduce the intractible factorisation of large prime products and integers to a (sub)polynomial time or trivial exercise on a normal PC in O(N). Working from the big end of any binary number, the carries of the previous smaller bits are already determined. This is VERY helpful... a solution to this factorisation intractibility is a BIG indicator that (counter-intuitively) P = NP and suggests we don't have our numbers organised in a 'logical' way...
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sirjon
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The title of this thread is "quick way to calculate the square root of large number. MSM 3 can be improved by using the following techniques...calande wrote:Hi guys,
Do you know how to calculate quickly √576 ?
What's your method?
Thanks
Ex. What is the sq, root of 14,965,872,048, an eleven digit number
Of course, the first thing to do is to group the given digits by twos... 1'49'65'87'20'48
First technique: Use of DOUBLE DOT NOTATION
The square value, nearest but less than 1'49.. (use a 'double dot' sign, to indicate that there are still digits after 1'49) is of course = 1'00...
20..^2 = 4'00.. (^2 means square, or in the power of 2)
15..^2 = 2'25..
10..^2 = 1'00..
Second technique: Use of 4DOWN ARROW or 5UP ARROW SIGNS
1'49.. is in between 2'25.. and 1'00.. . Write 4 and draw a downward arrow, to indicate that the next digit is either 4 or less (3, 2 or 1)
Use the add and divide method (taking the average of 1'00.. and 2'25..)
4/3 H = 2'25..
---- M =
2/1 L = 1'00..
ADD- = 3'25.. /2
AVER = 1'62.. - ..6 = 1'56
4/3 H = 2'25..
---- M = 1'56
2/1 L = 1'00..
1'49.. is in-between 1'56 and 1'00.
Encircle 2/1 and use the digits 2 and 1 as sample digits
12..^2 = 1'04..
)1 x ..4 = 04..
12..^2 = 1'44..
12..^2 = 1'44.. is applicable, since it is nearest but less than 1'49...
Third Technique: USe of A BOX notation
Enclose 12..^2 = 1'44.. inside a box, emphasizing that the value is important in dealing with the next step... (knowing the next middle square value)
125...^2 = 1'44'25..
12 x ..10 = -'12 '0..
125...^2 = 1'56'25... which is very large compared to the first digits of the given number (which is 1'49'65'..)
Use of HALF CIRCLES W/ DOWNWARD or UPWARD ARROWS
As a technique, after you taken the middle half square (example, 1'56'25), draw on its right side a half circle in downward arrow, meaning it is over and write on its left, a 4 w/downward arrow to indicate that the next digits is either 4 or below it.
Drawing a Half-circle in downward arrow - means 'over' then write 4down arrow
Drawing a Half-circle in upward arrow - means 'under' then write 5 up arrow
Take note: WE only use the double dots notation only up to the 'tens decimal place'. The decimal point is located after ..'48 (1'49'65'87'20'48), therefore, we should no longer use the double dot notation in this area and after.
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there is another way to calculate a square root of a perfect square.. the method is .. in above examplecalande wrote:Hi guys,
Do you know how to calculate quickly √576 ?
What's your method?
Thanks
first no. is 6.. tat means we have 2 option.. 6 and 4 ..
second.. forget about the 2nd no.. here 7
remains 5.. which will give us some the next result....that we will see whose square is near to the 5 .. that is 2 x2 = 4 so now we have 2 possibilities
24 and 26 .. now we can check
simillarly.. for finding we can find the square root of a no.. like.. 11236
s1- here 6 gives option 4 and 6
s2 leave 3
s3... nearest no. whose square is less than 112 is 10..
thus our result is 104 0r 106 and just with little practice u will get the result as 106 mentally
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sirjon
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biswajit1, the method you just discussed is based only (or in fact, effective only) for 3 or 4 digit given perfect square numbers only. In fact I saw one in youtube. As long as the last digit ends w/ 1, 4, 6, or 9, you can assume that the given problem (such as square root of 576) is indeed a perfect square since it ends w/ 6. The following can help you out:biswajit1 wrote:
...there is another way to calculate a square root of a perfect square.. the method is .. in above example
first no. is 6.. tat means we have 2 option.. 6 and 4 ..
second.. forget about the 2nd no.. here 7
remains 5.. which will give us some the next result....that we will see whose square is near to the 5 .. that is 2 x2 = 4 so now we have 2 possibilities
24 and 26 .. now we can check
simillarly.. for finding we can find the square root of a no.. like.. 11236
s1- here 6 gives option 4 and 6
s2 leave 3
s3... nearest no. whose square is less than 112 is 10..
thus our result is 104 0r 106 and just with little practice u will get the result as 106 mentally)[/spoiler]
square of 1 = 1 while the square of 9 = 81, both ends w/1 and their sum = 10 (1+9 =10)
sq. of 2 = 4 while sq. of 8 = 64, both ending in 4 and 2+8 = 10
sq. of 3 = 9 while sq. of 7 = 49, both ending in 9 and 3+7 = 10
sq. of 4 = 16 while sq, of 6 = 36, both ending in 6 and 4+6 = 10
Sometimes I call it the "ten and square" relationship
But my method, is indeed applicable to even a very large value (not only to a three digit number as 576) and even not in perfect square.
The idea of getting the middle square value (squares in the multiple of five) is to detect if a "certain digit w/in the square root" is either less than 5 or greater than 5. Then, the add and divide by two will minimize the choices into two digits only.
I believe, this is the easiest method, so far. Please call it MSM3, as a token of appreciation. Thanks a lot.
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- Amit Rawat
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- Stuart@KaplanGMAT
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Before this gets way out of hand, something important to note: the GMAT never requires you to do crazy calculations to solve questions. GMAC knows that, in real life, people have calculators (or computers, or accounting departments) to do hard math and you're not being tested on whether you're a human calculator.Amit Rawat wrote:what is shortcut of cube root
So, while it's often very useful to estimate roots of, at most, 3 digit numbers, you'll never have to actually find root(142543) or cuberoot(151234) or anything similar.
Also, remember that on the GMAT you have answer choices to help you out; if as a final step to solve a question you needed to find cuberoot(729), you could cube the choices to find a match (it's much easier to solve 9*9*9 than to try to calculate cube roots).
Here are two key cube root facts that could be relevant to the GMAT:
1) unlike square roots, cube roots have exactly 1 solution, which has the same sign as what you're cube rooting; and
2) perfect cubes are composed of triples of primes. For example, 1000 = 2^3 * 5^3. This fact can be very helpful in both data sufficiency and problem solving.

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sirjon
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I agree with you sir, the reason why, most of the time, I practice my MSM-3 with small numbers such, getting the square root of 3, 5, 8, etc... Although, I wish to point it out that, indeed, this is the simplest, direct and easiest way (or I might say, the quick way), so far, to calculate the square root of large number, as Beastly noted. ThanksStuart Kovinsky wrote:
So, while it's often very useful to estimate roots of, at most, 3 digit numbers, you'll never have to actually find root(142543)
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