When 'n' is a member of a Pythagorean
triplet, then the triplet is
n² - 1, 2n, n² + 1
or (13²- 1), (2 x 13), (13² + 1)
or (169 - 1), (26), (169 + 1)
or 168, 26 and 170

Since,S x 5 = 25
\(\therefore\) The unit digit of (55555i will be 5.

4, 9,16,25,36 and 49 are six perfect squares, which lie between 1 and 50.

Therefore, \(\sqrt { 7921 } \)= 89

We know that, the numbers end with 2, 3, 7 or 8, can never be a perfect square.
In a number 153, end digit is 3, which is one of 2, 7 or 8. So, it never be a perfect square.

86436

9

Using the following pattern for finding the square of the given number, we get
           1² = 1
          11²=121
       111² = 12321
      1111² = 1234321
    11111² = 123454321
  111111² = 12345654321
 1111111² = 1234567654321
1111111² = 1234567654321

158 is an even number, so its square is also even.

52

We have 

This shows that 22² < 525.
Next perfect square is 23² = 529
Hence, the number to be added is 23²- 525 = 529 - 525 = 4
Therefore, the perfect square so obtained is 525 + 4 = 529
Hence, \(\sqrt { 529 } \)= 23

We have 288 = 2 x 2 x 2 x 2 x 2 x 3 x3 = 22 X 22 X 2 X 32 = (2 x 2 x 3)2 X 2
\(\Rightarrow \frac { 288 }{ 2 } =\frac { (2\times 2\times 3)^{ 2 }3\times 2 }{ 2 } \)
\(\Rightarrow\)144 = (2 x 2 x 3)²
\(\Rightarrow\) \(\sqrt{144}\) = 2 x 2 x 3 = 12
Thus, 288 is to be divided by 2 to get a perfect square.
i.e., The required smallest number = 2
Also \(\sqrt{144}\) = 12.

We have

This shows that 57² is less than 3250 by 1. This means if we subtract the remainder (1) from the number, we This shows that 282 is less than 825 by 41. This means if we subtract the remainder (41) from the number, we get a perfect square. So, the required least number is 41.get a perfect square. So, the required least number is 1
Therefore, the required perfect square is 825 - 41 = 784
Hence,\(\sqrt { 784 } \)= 228

Given number is 180.
By using prime factorisation, we get
180 = 2 \(\times\) 2 \(\times\)3 \(\times\)3 \(\times\) 5
It is clear that in order to get a perfect square, one more 5 is required.
So, .the given number should be multiplied by 5 to make the product a perfect square.
\(\therefore\) 180 \(\times\) 5 = 900 is a perfect square.
Now, 900 = 2 \(\times\) 2 \(\times\)3 \(\times\)3 \(\times\) 5 \(\times\)5
=(2\(\times\)3\(\times\)5)²
\(\therefore\) \(\sqrt { 900 } \)= 2 \(\times\)3 \(\times\) 5 = 30
Hence, the square root of 900 is 30.

According to the first three pattern, if the first and the last digits of a given number is 1 and having n zeroes between it, then the square of that given number gives out the number whose first and last digits is 1, the middle digit is 2 and between the digits 1 and 2 and the digits 2 and 1, n zeroes digits exist.
In the number 100001², there are four zeroes between first and last digits.
So, in the missing digits 1..................2..................1, four zeroes lie between 1 and 2 and also four zeroes lie between 2 and 1.
Hence, the required number is 10000200001.
In the number, 10000001², there are six zeroes between first and last digits.
So, the required number is 100000020000001.

1.SQUARE
2.PYTHAGOREAN
3.PERFECT SQUARE
4.TRIPLET
5.SQUARE ROOT
6.FACTOR

Resolving 256 into prime factors, 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Grouping the factors in pairs in such a way that both the factors in each pair are same.
256 = 2² x 2² x 2² x 2²
Here, 256 can be grouped into pairs of equal factors and no factor is left over.

Hence, 256 is a square number.
Again, 2 X 2 x 2 X 2 = 16
50,256 is the square of 16.

To find the square roots of 100, we subtract successive odd number starting from 1 as follows:
100 - 1 = 99, 99 - 3 = 96, 96 - 5 = 91, 91 - 7 = 84
84 -9 = 75, 75 -11 =64, 64 -13 = 51,51-15 =36
36-17 = 19, 19 - 19 = 0
We observe that the number 10 reduced to zero after subtracting first 10 odd numbers.
So, 100 is a perfect square.
\(\therefore \ \sqrt { 100 } =10\)
Hence, the square root of 100 is 10.

\(\sqrt{10000}=100\)

54 x 6 = 324 = 18²

7 x 7 = 49

2 x 2 = 4

2 x 12 = 24

Number of zeros at the end of the number 9000 = 3
∴ Number of zeros at the end of the square of the number 9000 = 2 x 3 = 6.

None of 51, 52, .........., 59 is a perfect square.

Let the required number be x.
Then, according to the question
x³ = x ...(1) and x²≠x ...(2)
From (1), x³ -x = 0
⇒ x(x² - 1) = 0
⇒ x = 0, ± 1
⇒ x = 0,1,-1
If x = 0, then x² = x.
∴ x = is inadmissible
If x = 1 then x² = x.
∴ x = 1 is inadmissible
If x = - 1, then x² = (- 1)2 = 1 ≠ x(= - 1)
Hence, the required number is - 1.

Let the number of plates on the table be y and the number of puris be x.
Then, according to the first condition of the problem
4(y - 1) = x ...(1)
and, according to the second condition of the
problem
3y + 1= x    ... (2)
From (1) and (2), we have
4(y - 1) = 3y + 1
⇒ 4y - 4 = 3y + 1
⇒ 4y - 3y = 1 + 4
⇒ y=5
Put y = 5 in (1), we get
x = 4(5) - 4 = 20 - 4 = 16
Hence, the number of plates and number of puris on the table are 5 and 16 respectively.