Squaring Numbers Ending in "5"
In Vedic Mathematics, squaring numbers that end in 5 can be done very quickly using a simple and effective formula. This method significantly reduces the time and effort compared to traditional methods.
Below is the step-by-step process.
Step 1: Add 1 to the first digit from the left and multiply by the number itself.
Step 2: Add 52 = (25) at the end to the number obtained from step 1.
Find 252
To get the first part, multiply the tens digit, 2, by the next higher number, which is 3.
2 x 3 = 6
The second part is always 25.
So, 252 = 625.
Find 352
To get the first part, multiply the tens digit, 3, by the next higher number, which is 4.
3 x 4 = 12
The second part is always 25.
So, 352 = 1,225.
Find 552
To get the first part, multiply the tens digit, 5, by the next higher number, which is 6.
5 x 6 = 30
The second part is always 25.
So, 552 = 3,025.
Find 652
To get the first part, multiply the tens digit, 6, by the next higher number, which is 7.
6 x 7 = 42
The second part is always 25.
So, 652 = 4,225.
Find 752
To get the first part, multiply the tens digit, 7, by the next higher number, which is 8.
7 x 8 = 56
The second part is always 25.
So, 752 = 5,625.
Find 952
To get the first part, multiply the tens digit, 9, by the next higher number, which is 10.
9 x 10 = 90
The second part is always 25.
So, 952 = 9,025.
Find 1052
To get the first part, multiply the tens digit, 10, by the next higher number, which is 11.
10 x 11 = 110
The second part is always 25.
So, 1052 = 11,025.
Algebraic Proof: (ax + b)2
Consider the identity: (ax + b)2 ≡ a2x2 + 2abx + b2
For x = 10 and b = 5, it becomes:
(10a + 5)2 = a2 × 102 + 2 × 10a × 5 + 52
= a2 × 102 + a × 102 + 52
= (a2 + a) × 102 + 25
= a × (a + 1) × 102 + 25
Clearly, 10a + 5 represents two-digit numbers such as 15, 25, 35, 45, ..., 95.
