**Submitted by Sav2718:**

I saw that someone wrote as a comment on my business card:

“I saw something similar for fifth roots- getting the last digit right is a lot easier too!”.

So I just wanted to add a word about the rightmost digit of a perfect root in general:

The pattern for the rightmost digit is the same for every order of the form 4K+1 (the digit itself) and for 4K-1 (the same unique pattern as with cube roots). the same is true for roots of an even-order but an extra step of elimination is needed, two steps for even-order of the form 4K.

For example, lets take a look at some (perfect) square roots:

1^2=1

2^2=4

3^2=9

4^2=16

5^2=25

6^2=36

7^2=49

8^2=64

9^2=81

10^2=100

11^2=121

12^2=144

13^2=169

14^2=196

15^2=225

16^2=256

17^2=289

18^2=324

19^2=361

20^2=400

etc.

the pattern for the rightmost digit is always (0,)1,4,9,6,5,6,9,4,1(,0)

1 for 1 and 9 (the complement to 10)

4 for 2 and 8

9 for 3 and 7

6 for 4 and 6

5 for 5

a trail of 2N 0s (even number) in the right hand of the number beneath the root sign will be a trail of N 0s in right hand of the answer (ex: sqrt(15210000)=3900).

Now in order to calculate a given perfect square root lets look at an example:

sqrt(5329)

70^2(=4900)<5329<80^2(=6400)

or even more simplified:

7^2(=49)<53<8^2(=64)

so the leftmost digit of the answer is 7

5329 ends with a 9 so according to our pattern the rightmost digit is either 3 or 7 but how can we tell?

Let’s have a look at the trick for squaring a number that ends with 5:

(10a+5)^2=100a^2+2*10*5*a+5^2=100a^2+100a+25=100a*(a+1)+25

or even more simplified:

[a*(a+1)]&[25] (where “&” is the concatenation operator)

examples:

35^2=[3*4]&[25]=[12]&[25]=1225

75^2=[7*8]&[25]=[56]&[25]=5625

435^2=[43*44]&[25]=[1892]&[25]=189225

you may also prefer to calculate it as such:

35^2=[3^2+3]&[25]=[9+3]&[25]=[12]&[25]=1225

75^2=[7^2+7]&[25]=[49+7]&[25]=[56]&[25]=5625

435^2=[43^2+43]&[25]=[1849+43]&[25]=[1892]&[25]=189225

Now back to our square root:

we know sqrt(5329) is either 73 or 77 but we can quickly check that

75^2=[7*8]&[25]=[56]&[25]=5625>5329

so if 75 is too big, 77 is even bigger than the answer must be (correctly) 73.

Note that because a square of number ending with 5 always ends with 25 one doesn’t even need to calculate the whole thing and just have a look at 7*8 (7 times its successor).

Lets do another one quickly:

sqrt(107584)

32^2(=1024)<1075 (don’t need the ‘84’) <33^2(=1089)

32_

107584 ends with 4 so the rightmost digit is either 2 or 8

32^2+32=1024+32=1056<1075

too small so 322 is even smaller

so the answer is (correctly) 328.

As I already mentioned even-order roots from the form 4K need another step of elimination.

Higher numbers require more steps and non-perfect roots require algorithms from a different kind and approach but I am willing to write about those in the future.

He’s referring to **this** awesome submission.