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Recently I read this, which teaches a trick for finding the square root of a number very quickly and was quite astounded. How/why does this method work?

For those of you with linkphobia, I copied and pasted part of the page below:

Step-1: Look at the magnitude of the “hundreds number” (the numbers preceding the last two digits) and find the largest square that is equal to or less than the number. This is the 1st part of the answer.

Step-2: Now, look at the last (unit’s) digit of the number. If the number ends in a:

$0 \rightarrow$ then the ending digit of the answer is a $0$.
$1 \rightarrow$ then the ending digit of the answer is $1$ or $9$.
$4 \rightarrow$ then the ending digit of the answer is $2$ or $8$.
$5 \rightarrow$ then the ending digit of the answer is a $5$.
$6 \rightarrow$ then the ending digit of the answer is $4$ or $6$.
$9 \rightarrow$ then the ending digit of the answer is $3$ or $7$.

To determine the right answer from $2$ possible answers (other than $0$ and $5$), mentally multiply the findings in step-1 with its next higher number. If the left extremities (the numbers preceding the last two digits) are greater than the product, the right digit would be the greater option $(9,8,7,6)$ and if left extremities are less than the product, the right digit would be the smaller option $(1,2,3,4)$.

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  • $\begingroup$ It'It's important to note this only works on numbers that are perfect squares. It will not allow you to estimate an irrational square root. It's easy to get the square root of something like 44xy must be be 60 and 70 because 44 is between 6^2 and 7^2. So it's just a matter of guessing the second digit. If we are given that 4489 is a square, we know the second digit must be 3 or 7. 3 will give as 3600 + 60x3x2 + 9 approx 3600 + 400 is way too but 7 gives us 3600 + 60x7x2+49 approx 3600 + 800 or 900 which is just about right. So it must be 67. But ... $\endgroup$ – fleablood Dec 23 '16 at 8:18
  • $\begingroup$ If we we asked to find the square root of 4365 we'd be sol. We'd know it must be between 60 and 70. But to find the second digit we'd assume it has to be 5, so it must be 65. But it isn't ... because there is no integer square root at all. $\endgroup$ – fleablood Dec 23 '16 at 8:21
  • $\begingroup$ @hardmath Added a bit $\endgroup$ – suomynonA Dec 23 '16 at 16:55
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First, it's important to note that this only works for roots of perfect squares.

Start by examining what happens to multiples of 10 when you square them: $$10^{2}=100$$ $$20^{2}=400$$ $$30^{2}=900$$ $$40^{2}=1600$$ $$50^{2}=2500$$ $$60^{2}=3600$$ $$70^{2}=4900$$ $$80^{2}=6400$$ $$90^{2}=8100$$ So, because they're multiples of 10, the squares all wind up being multiples of 100. Yes, adding the numbers 1 through 9 will increase this, but even, say, $79^{2}$ can't be greater than $80^{2}$. Therefore, if a perfect square is between 4900 and 6400, then it's root must be 70-something.

That explains getting the tens digit, but what about the units digit?

Look at the patterns of the squares of the one-digit numbers, paying particular attention to the units digit:

$$1^{2}=\textbf{1}$$ $$2^{2}=\textbf{4}$$ $$3^{2}=\textbf{9}$$ $$4^{2}=1\textbf{6}$$ $$5^{2}=2\textbf{5}$$ $$6^{2}=3\textbf{6}$$ $$7^{2}=4\textbf{9}$$ $$8^{2}=6\textbf{4}$$ $$9^{2}=8\textbf{1}$$

Notice the units digits of the squares go up 1, 4, 9, 6, 5, and then go back down in reverse order, 6, 9, 4, 1.

So, a perfect square ending in 1, must have a root ending in 1 or 9. A perfect square ending in 4 must have a root ending in 2 or 8, and so on.

For the perfect square of a 2-digit number, you've already narrowed it to 2 possibilities.

The puzzling part about this is probably how you choose between the 2. The standard method is to learn a quick trick for squaring numbers ending in 5.

To square a 2-digit number ending in 5, you take the tens digit, multiply it by 1 greater than itself, and put 25 to the left of that result. For 25, then tens digit is 2, so we multiply 2 by 3 to get 6, stick a 25 on the end, and we have 625. To square 35, we'd multiply 3 (the tens digit) by 4 (one more than the tens digit) to get 12, add the 25 to the end, giving us 1225.

For use in determining roots, wait until you get to the part where you have 2 choices. For example, given 784, you can work out that the tens digit of the answer is in the 20s, and that the units digit must be 2 or 8. How do you decide?

Well, using the trick you just learned, you can quickly calculate that 25 squared is 625. Is 784 more or less than 625? Obviously, it's more, which means the square root must be the greater of the 2 choices, so the square root of 784 must be 28.

The article to which you link tries a bit of a short cut by not teaching you the full square-a-number-ending-in-5 trick, but in my opinion, it just seems to make the whole trick more confusing.


Why does the square-a-number-ending-in-5 trick work?

Looking at this algebraically, you can think of squaring a number ending in 5 as: $$(10x+5)^2=(10x+5)(10x+5)$$ This multiplies out to: $$100x^{2}+100x+25$$ We can factor this as: $$100(x^2+x)+25$$ And we can even factor it once more: $$100\{x(x+1)\}+25$$ There's the trick itself laid bare. Multiply x by 1 greater than itself, multiply it by 100 to move the answer over 2 places to the left, and then add 25.

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Write $n = 10x + y$ where $y \in \{0, 1, \dots, 9\}$. Then

$n^2 = 100x^2 + 20xy + y^2$

Note that the "magnitude of the hundreds numbers" of $n^2$ (as in the link) is $x^2 + 0.2xy$ if $xy \geq 5$ and $x^2$ otherwise. Since $y < 10$, we may characterize $x$ as the largest integer smaller or equal to the square root of the "magnitude of the hundreds number". That's step 1 in the link.

The congruence $y^2 \equiv y \mod 10$ has two solutions. Those are the options for the last digit in step 2.

Finally note that $(x + 1)x$ is $x^2 + x$. The "magnitude of the hundreds" is bigger than $x^2 + x$ if and only $y \geq 5$. This allows you to choose one and only one solution from step 2. That's what the last step in the method is saying.

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