In this post we will discuss some important properties of square of numbers with example.
For reference, given below is the table of numbers and its perfect squares.
| Number | Perfect Square |
|---|---|
| 1 | 1 |
| 2 | 2 |
| 3 | 9 |
| 4 | 16 |
| 5 | 25 |
| 6 | 36 |
| 7 | 49 |
| 8 | 64 |
| 9 | 81 |
| 10 | 100 |
| 11 | 121 |
| 12 | 144 |
| 13 | 169 |
| 14 | 196 |
| 15 | 225 |
| 16 | 256 |
| 17 | 289 |
| 18 | 324 |
| 19 | 361 |
| 20 | 400 |
| 21 | 441 |
| 22 | 484 |
| 23 | 529 |
| 24 | 576 |
| 25 | 625 |
| 26 | 676 |
| 27 | 729 |
| 28 | 784 |
| 29 | 841 |
| 30 | 900 |
| 31 | 961 |
| 32 | 1024 |
| 33 | 1089 |
| 34 | 1156 |
| 35 | 1225 |
| 36 | 1296 |
| 37 | 1369 |
| 38 | 1444 |
| 39 | 1521 |
| 40 | 1600 |
| 41 | 1681 |
| 42 | 1764 |
| 43 | 1849 |
| 44 | 1936 |
| 45 | 2025 |
| 46 | 2116 |
| 47 | 2209 |
| 48 | 2304 |
| 49 | 2401 |
| 50 | 2500 |
Properties of Square numbers
Given below is the description of properties with examples.
Property 01
A square number always end with digit 0, 1, 4, 5, 6, 9.
For example;
Numbers like 100, 81, 144, 25, 36, 49 are all examples of perfect square.
So in competition exam, if you are asked to check if the given number is perfect square or not, first check if the given number is ending with 0, 1, 4, 5, 6, 9 or not.
Example 01
Identify 92 is perfect square or not
Solution
No need to do any analysis.
The given number is ending with 2, hence it cannot be a perfect square.
Example 02
Check if the number 733 is perfect square or not?
Solution
The number is ending with 3, hence it cannot be a perfect square.
Knowing this property can save lot of valuable time in examination. So make sure to remember this important point.
Property 02
Number ending with odd number of zero is never a perfect square.
For example:
7000 ⟹ three zeros at the end
3100000 ⟹ five zeros at the end
90 ⟹ one zero at the end
All of the above numbers have odd number of zeros at the end. Hence, these numbers are not perfect square.
Property 03
Square of even number is always an even number.
Similarly square of odd number is always an odd number.
Consider the square of even numbers.
\mathtt{8^{2} =64}\\\ \\ \mathtt{12^{2} =144}\\\ \\ \mathtt{20^{2} =\ 400}
Note that on squaring, we have got all even numbers.
Now square the odd numbers.
\mathtt{3^{2} =9}\\\ \\ \mathtt{9^{2} =81}\\\ \\ \mathtt{15^{2} =\ 225}
Note that on squaring the odd numbers we get the odd numbers.
Property 04
Last digit of number and the square.
(i) Number with end digit 1 & 9 have perfect square with end digit 1.
For example;
\mathtt{11^{2} =121}\\\ \\ \mathtt{21^{2} =441}\\\ \\ \mathtt{9^{2} =\ 81}\\\ \\ \mathtt{19^{2} =\ 361}
(ii) Number with end digit 2 & 8 have perfect square with end digit 4.
\mathtt{12^{2} =144}\\\ \\ \mathtt{32^{2} =1024}\\\ \\ \mathtt{8^{2} =\ 64}\\\ \\ \mathtt{18^{2} =\ 324}
(iii) Number with end digit 3 & 7 have perfect square with end digit 9.
\mathtt{13^{2} =169}\\\ \\ \mathtt{23^{2} =529}\\\ \\ \mathtt{7^{2} =\ 49}\\\ \\ \mathtt{17^{2} =\ 289}
(iv) Number ending with 4 and 8 have perfect square with end digit 6.
\mathtt{4^{2} =16}\\\ \\ \mathtt{14^{2} =196}\\\ \\ \mathtt{6^{2} =\ 36}\\\ \\ \mathtt{26^{2} =\ 676}
Property 05
Shortcut to find last digit of perfect square
The last digit of perfect square is equal to the square of unit place of original number.
Example 01
Find the last digit of square of number 48.
Here the unit place digit is 8. Square this number.
\mathtt{8^{2} =\ 64}
Now last digit is 4.
Hence, the last digit of perfect square of number 48 is 4.
\mathtt{48^{2} =\ 2304}
Example 02
Find the last digit of square of number 63.
Solution
In number 63, unit place is 3.
Squaring the unit place.
\mathtt{3^{2} =\ 9}
Hence, 9 will be the last digit of the perfect square.
Let’s check by squaring the number 63.
\mathtt{63^{2} =\ 3969}
Note that the last digit is 9.
Hence Proved.
Property 06
The square of proper fraction is always smaller than original fraction
The fraction in which numerator is less than denominator is called proper fraction.
Let us understand the property with example.
Example 01
Consider the fraction 4/5
On squaring the fraction we get;
\mathtt{\left(\frac{4}{5}\right)^{2} =\ \frac{16}{25}}
Given below is the value of both the fraction.
4/5 = 0.8
16/25 = 0.64
So, 16/25 < 4/5
Hence, the square of proper fraction is less than original number.
Property 07
Value of square of number is equal to sum of consecutive odd number
For example; consider the number \mathtt{5^{2}}
The value of \mathtt{5^{2}} is equal to sum of 5 consecutive odd number.
\mathtt{5^{2}} = 25
Value of 5 consecutive odd number is also 25.
1+ 3 + 5 + 7 + 9 = 25
Example 02
Consider the number \mathtt{7^{2}}
The value of \mathtt{7^{2}} is equal to sum of 7 consecutive odd number.
Value of square
\mathtt{7^{2}} = 49
Value of sum of 7 consecutive odd number.
1+ 3 + 5 + 7 + 9 + 11 + 13 = 49.
Property 08
If three numbers are in the form of \mathtt{2m,\ m^{2} -1\ \&\ m^{2} +1} , then they are part of Pythagoras Triplets.
Where, m is a natural number greater than 1.
Let us understand the concept with example.
Example 01
Let m = 2.
Then value of three numbers are;
2m = 2 x 2 = 4
\mathtt{m^{2} -1\ \Longrightarrow \ 2^{2} -1=\ 3}\\\ \\ \mathtt{m^{2} +1\ \Longrightarrow \ 2^{2} +1\ =\ 5}
Here all the three number 3, 4 and 5 form Pythagoras triplets.
Example 02
Let m = 3;
2m = 2 x 3 = 6
\mathtt{m^{2} -1\ \Longrightarrow \ 3^{2} -1=\ 8}\\\ \\ \mathtt{m^{2} +1\ \Longrightarrow \ 3^{2} +1\ =\ 10}
Here the numbers 6, 8 and 10 form Pythagoras triplets.
Perfect Square Property – Solved examples
(01) Check if the below number are perfect square or not.
(i) 23
(ii) 16
(iii) 37
(iv) 144
(v) 52
(vi) 1225
(vii) 1000
Solution
(i) 23
Not a perfect square.
Perfect square doesn’t contain number 3 in unit place.
(ii) 16
It is the square of number 4.
(iii) 37
Not a perfect square.
As per property 01, no perfect square contain 7 as last digit.
(iv) 144
It is the perfect square of number 12.
(v) 52
Not a perfect square.
No perfect square contain 2 at the unit place.
(vi) 1225
Perfect square of number 35.
(vii) 1000
Not a perfect.
No perfect square contain odd number of zeros at the end.
(02) Find the last digit of perfect square of given numbers.
(i) 35
(ii) 108
(iii) 67
(iv) 93
(v) 100
Solution
Last digit of perfect square can be found by taking square of unit digit of original number and selecting its last digit.
(i) 35
Unit digit is 5.
Taking square of unit digit.
\mathtt{5^{2} =25}
Here again unit digit is 5, so the last digit of perfect square is also 5.
(ii) 108
Here unit digit is 8.
Taking square of unit digit.
\mathtt{8^{2} =64}
Here unit digit is 4.
So, last digit of square of 108 is 4.
(iii) 67
Here unit place is 7.
Taking square of unit digit.
\mathtt{7^{2} =49}
Here unit place is 9.
So, last digit of square of 67 is 9.
(iv) 93
Here unit place is 3.
Taking square of unit digit.
\mathtt{3^{2} =9}
Here unit place is 9.
So, last digit of square of 93 is 9.
(v) 100
Last digit is 0.