Squares of numbers ending in 5

Vedic Mathematics provides us with simple ways to solve math problems. For instance, if you are looking to square numbers ending in ’5′, it can be done using the Ekadhikena Purvena sutra.

Roughly translated, Ekadhikena Purvena means:

“By one more than the previous one.”

Example #1: 35^2 or “35 * 35″
1. 35 can be broken down into ’3′ and ’5′.
2. Square the ’5′, and write it down. (25)
3. Looking at the sutra, we notice that we have to identify what the ‘previous one’ is. In our case ’3′ is the previous one.
4. Now we multiply ’3′ by one more than the previous one, so 3 * 4 = 12.
5. Combine the two numbers ’12′ and ’25′, and we get our answer. 1225

Example #2: 75^2
1. 75 can be broken down into ’7′ and ’5′
2. 5^2 = 25.
3. Identify ‘previous one’, which is ’7′ for us.
4. 7 * (7 + 1) = 56
5. Combine the two. 5625

Now to ensure that this is always valid we must prove it algebraically.

Two Digit Square ending in 5
For this, we’ll write out our number to square in the form of (ax + b), where x = 10, b = 5, and ‘a’ is some integer from 0 to 9.

Foil (ax + b) ^2;
(ax + b) ^2 = (a^2 * x^2) + 2axb + b^2

Substitute in x= 10 and b = 5;
(a*10 + 5) ^2 = (a^2)(10^2) + a(10^2) + 25

Factor out a 10^2;
(a*10 + 5) ^2 = (a^2 + a)(10^2) + 25

Now factor out an ‘a’;
(a*10 + 5) ^2 = (a)(a + 1)(10^2) + 25

Voila! As you can see, the (a)(a +1) corresponds to the “previous” times “previous + 1″, and the ’25′ is tacked onto the end. “By one more than the previous one” is also valid for numbers involving more digits, however I’ll leave the proof for you to do.

Practice Problems:
See how fast you can do these problems…
1. 55^2
2. 85^2
3. 105^2
4. 125^2

If you have any questions, let me know.