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You can also add the digits of the numbers until you get one digit like if you add the digits of 13 you get 4. If you take the square of 13, it has to add to the same amount as 4x4. 4x4 is 16. 1+6 is 7. So the digits of the square of 13 also have to add to 7. The square of 13 is 169. Add the digits: 1+6+9=16, 1+6=7. So if you can’t tell by looking which is closer, add the digits of the numbers until you get one digit. Take the square and add the digits until you get one number. Look at your two choices and add their digits until you get one number and see which one matches. Let’s say you’re deciding between 43 and 47. The number that is the square of one of these numbers is 2209. 2+2+0+9=13, 1+3=4. What’s the square root of 4? 2. So when you add the digits of the square root of 2209, it has to add to 2: 4+3=7 so it can’t be 43. 4+7=11, 1+1=2. It’s definitely 47. The only time this doesn’t work is when you are choosing between multiples of 3. For example, 42 and 48. The digits of both their squares will add to 9. So you have to guess. These one digit numbers that all numbers add down to are called digital roots. You can add the digits as I did or you can enter a number on a scientific calculator and hit mod 9 to get the digital root. These digits are also the remainders when you divide by 9. Adding them and multiplying them is part of modular arithmetic.

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The big problem with your method is that it only works if you are sure that the given root has an integer solution. If you don't know this, you should use the approximation formula Sqrt(n² + a) ≈ n + a/(2n) and (n + 0.5)² = n * (n+1) + 0.25 for better squares between the squares of natural numbers. *RULESET* Given: Sqrt(x) *(1) Remove redundant powers of 10* If x > 1000: 10 * Sqrt(x/100) If x > 100,000: 100 * Sqrt(x/10,000) If x > 10 * 10^(2k): 10^k * Sqrt(x/10^(2k)) New Radikand: y = x/10^(2k) Advantage: 10 < y < 1000 You only need Square numbers of 1 to 31. *(2) Find the closest square number n* Bonus: If the number is relatively centered between two square numbers n1 and n2, you can form the product of these and add 0.25 (n1 * n2 + 0.25) to use the square number (n1 + 0.5)². *(3) Calculate the difference* a = y - n² *(4) Use approximation formula* Sqrt(y) = Sqrt(n² + a) ≈ n + a/(2n) ------------------------------------- Examples: Sqrt(1156) = Sqrt(11.56 * 100) = 10 * Sqrt(11.56) = 10 * Sqrt(12.25 - 0.69) = 10 * Sqrt(3.5² - 0.69) ≈ 10 * (3.5 - 0.69/(2*3.5)) ≈ 10 * 3.4 = 34 Sqrt(2304) = Sqrt(23.04 * 100) = 10 * Sqrt(23.04) = 10 * Sqrt(25 - 1.96) = 10 * Sqrt(5² - 1.96) ≈ 10 * (5 - 1.96/(2*5)) = 10 * 4.8 = 48 Sqrt(4489) = Sqrt(44.89 * 100) = 10 * Sqrt(44.89) = 10 * Sqrt(42.25 + 2.64) = 10 * Sqrt(6.5² + 2.64) ≈ 10 * (6.5 + 2.64/(2*6.5)) ≈ 10 * 6.7 = 67 Sqrt(12996) = Sqrt(129.96 * 100) = 10 * Sqrt(129.96) = 10 * Sqrt(132.25 - 2.29) = 10 * Sqrt(11.5² - 2.29) ≈ 10 * (11.5 - 2.29/(2*11.5)) ≈ 10 * 11.4 = 114 Sqrt(24649) = Sqrt(246.49 * 100) = 10 * Sqrt(246.49) = 10 * Sqrt(240.25 + 6.24) = 10 * Sqrt(15.5² + 6.24) ≈ 10 * (15.5 + 6.24/(2*15.5)) ≈ 10 * 15.7 = 157 And for error estimation for the approximation formula you can use the following term: Error = a²/(8n³) With a < n: Error < n²/(8n³) = 1/(8n) ------------------------------------- Without the 0.5 formula: Sqrt(1156) = Sqrt(11.56 * 100) = 10 * Sqrt(11.56) = 10 * Sqrt(9 + 2.56) = 10 * (3² + 2.56) ≈ 10 * (3 + 2.56/(2*3)) ≈ 10 * 3.4 = 34 Sqrt(4489) = Sqrt(44.89 * 100) = 10 * Sqrt(44.89) = 10 * Sqrt(49 - 4.11) = 10 * Sqrt(7² - 4.11) ≈ 10 * (7 - 4.11/(2*7)) ≈ 10 * 6.7 = 67 Sqrt(12996) = Sqrt(129.96 * 100) = 10 * Sqrt(129.96) = 10 * Sqrt(121 + 8.96) = 10 * Sqrt(11² + 8.96) ≈ 10 * (11 + 8.96/(2*11)) ≈ 10 * 11.4 = 114 Sqrt(24649) = Sqrt(246.49 * 100) = 10 * Sqrt(246.49) = 10 * Sqrt(256 - 9.51) = 10 * Sqrt(16² - 9.51) ≈ 10 * (16 - 9.51/(2*16)) ≈ 10 * 15.7 = 157

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Lovely teaching! Also instead of doing: For example on the square root of 2304: its either 42 or 48 and instead of finding the square root of 40 and 50, you can find the square root of 45 which is in the middle of both numbers. An easy way to work out how to square numbers ending in 5: 5 x 5 is *25,* then keep that. for 45, you have done the 5 and you have 4 left. Add one to that number then times both numbers: 4 times 5 which gives you *20.* Put the numbers together in the order they were in: 4 5 20 25 Your answer is 2025 The original number was 2304 (The number that you had to work out the square root of) which is higher than 2025 so the answer has to be higher too. The answer is 48. Sorry If I didnt explain well, I just think this is a helpful way to do it. It is a bit faster. :)

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An easier way to find the unit digit is to multiply the 10s digit by its next consecutive number. That is 3*4=12. Since 12 is larger than 11 you take the smaller digit 3. If 11 was larger than this product you would take the larger 6.

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MR. Organic Chemistry Tutor, thank you for another monster video/lecture on How to find the square root of large numbers mentally. This topic shows some mathematical tricks that all students should know in Mathematics. There are different methods for computing the square root of large numbers.

There’s another way to do it, too. After determining the two possible digits in the ones column, ignore the last two digits, and find the nearest square below the remaining numbers and find the root. That’s the number for the tens column. Multiply that number by the next number up. If the product is smaller than the number left of the ignored digits, pick the larger number for the ones column.

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There was a mistake in the 15^2 it is actually 225 for people who are confused or thought they remembered wrong but very good method overall!thanks for sharing

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Let's use your last example - 24649 Now last digit or we say the digit at unit place is 9 so unit digit of sqrt. Of the given no. Should be either 3 or 7, now the nearest square of a no. is 225 which is equal or smaller than 246, so we take the sqrt of 225, i.e 15 now the desired answer could be either 153 or 157 for that we have to just multiply 15 by 16 i.e 240 (now we will check if 240 is smaller or equal to 246) 240

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There is also another method mentioned by Presh Talwalkar While finding a root of a number by this method, you may get 2 numbers to decide on. You can take the sqaure of the middle number which would be tge 5th term For example: For root of 24649 You have 2 options: 153 or 157 Now you can take the sqaure of the middle term which would be 155. There is a brilliant trick to find the square of a number ending in 5 Suppose 45 sqaured. Now leave the last digit which would be 5, so you have 4. Now multiply 4 with the number after itself, so: 4*5 which is 20. Now add 25 as the last 2 digits of the number so 45 squared is 2025

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This totally works for any digit number, you just need to expand the lines. Each starts one more digit to the right. The first line is digits right next to each other, the next is numbers one apart, the next 2 apart and so on until the last line is the first digit times the last digit. With a little practice, I was squaring a 6 digit number in under a minute.

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A useful trick is that a square of #5 number is # x (# + 1) +25, eg 65 squared is 4225 so you quickly see if you go up or down.

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This video is awesome. I've been asking how to do squaring without a calc without doing 2x2x2x2x2.... etc that was shown in my examples...