(1). The given number is first arranged in two-digit groups from right to left. If on left hand side, a single digit is left, that will also be counted as a group.

(2). The number of digits in the square root will be the same as the number of groups derived from the number. Examples are:

- 25 will be having one group as '25', hence square root should be of one digit.
- 144 will be having two groups as '44' and '1', hence the square root should be of two digits.
- 1024 will be having two groups as '24' and '10', hence the square root should be of two digits.

(4). The squares of the first nine natural numbers are 1,4,9,16,25,36,49,64, and 81. All of these squares end with 1, 4, 5, 6, 9, 0. This means

- An exact square never ends in 2, 3, 7 or 8
- If a number ends in 2, 3, 7 or 8, its square root will always be an irrational number
- If an exact square ends in 1, its square root ends in 1 or 9
- If an exact square ends in 4, its square root ends in 2 or 8
- If an exact square ends in 5, its square root ends in 5
- If an exact square ends in 6, its square root ends in 4 or 6
- If an exact square ends in 9, its square root ends in 3 or 7

(6). If a perfect square is an even number, the square root is also an even number

(7). A whole number, which ends with an odd numbers of 0's, can never be the square of a whole number

(8). An exact square never ends in a 6 if the penultimate digit(digit that is next to the last digit) is even (eg. exact squares can not end in 26, 46, 86, etc.)

(9).An exact square never has an odd penultimate digit unless the final digit is a 6 (thus, exact squares can not end in 39,71, etc.)

(10).An exact square never ends with an even number when the last two digits taken together are not divisible by 4 (thus, no exact square can end in 22, 34 and other non-multiples of 4 if the last digit is even)

**Firstly**, we use "

**" technique to solve the square root.**

*The First by the First and the Last by the Last***(1). √6889**

There are two groups of figures, '68' and '89'. So we expect 2-digit answer.

Now see since 68 is greater than 64(8

^{2}) and less than 81(9

^{2}), the first figure must be 8.

So, 6889 is between 6400 and 8100, that means, between 80

^{2}and 90

^{2}.

Now look at the last figure of 6889, which is 9.

Squaring of numbers 3 and 7 ends with 9.

So, either the answer is 83 or 87.

There are two easy ways of deciding. One is to use the digit sums.

If 87

^{2}= 6889

Then converting to digit sums

(L.H.S. is 8+7 = 15 -> 1+5 -> 6 and R.H.S. is 6+8+8+9 -> 31 -> 3+1 -> 4)

We get 6

^{2}-> 4, which is not correct.

But 83

^{2}= 6889 becomes 2

^{2}-> 4, so the answer must be 83.

The other method is to recall that since 85

^{2}= 7225 and 6889 is below this. 6889 must be below 85. So it must be 83.

**Note:**

*To find the square root of a perfect 4-digit square number we find the first figure by looking at the first figures and we find two possible last figures by looking at the last figure. We then decide which is correct either by considering the digit sums or by considering the square of their mean.***(2). √5776**

The first 2-digit(i.e. 57) at the beginning is between 49 and 64, so the first figure must be 7.

The last digit (i.e. 6) at the end tells us the square root ends in 4 or 6.

So the answer is 74 or 76.

74

^{2}= 5776 becomes 2

^{2}-> 7 which is not true in terms of digit sums, so 74 is not the answer.

76

^{2}= 5776 becomes 4

^{2}> 16 -> 7, which is true, so 76 is the answer.

Alternatively to choose between 74 and 76 we note that 75

^{2}= 5625 and 5776 is greater than this so the square root must be greater than 75. So it must be 76.

Second technique is useful for bigger numbers and in this method, we use "Duplex". In the next article, we shall continue to discuss this second technique. Until then, good luck and happy computing!!

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