Guest Post: Vedic Mathematics 2 – Squaring Numbers Ending in 5

Sanjay Guilati (author) has been teaching computer and mathematics in Bhilai , state Chhattisgarh in India for 15 years. Currently he is a  teacher in a seniorsecondary school and he is also involved in teacher training.  Sanjay is the writer of the new blog titled Mathematics Academy.

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Vedic Mathematics 2 – Squaring Numbers Ending in 5
by Sanjay Gulati

In Vedic Mathematics, there are two types of techniques: specific techniques and general techniques. The specific techniques are fast and effective but can only be applied to a particular combination of numbers. On the other hand,  Vertically and Crosswise method is one of the  general techniques in Vedic Mathematics , as it can be used to multiply numbers of any possible combination of digits. Thus general techniques have a much wider scope of application than specific techniques because they deal with a wider range of numbers. Here first we will discuss some of the specific techniques.

Square Numbers Ending in 5

By way of notation, 22 means 2 times 2 or 4 and 32 means 3 times 3 or 9. What if you are asked to calculate 452 or 652 or the square of any number ending in a 5 ( a number with unit digit 5) ? More than likely, we would need pencil and paper or a calculator to work it out. One of the Vedic Mathematics Sutras “one more than the one before” can help us to solve the questions immediately. Very simply, multiply the digit(s) to the left of the 5 by the next higher number and then write 25 after it.

For example, let’s apply this sutra to the calculation of 452.

1st step: Determine the number to the left of the 5. That number is 4 in our example.

2nd step: Multiply this number by next higher number . This means multiply the 4 by 5. This results in the number 20.

3rd step: Follow this result with the number 25. This means the number 25 will follow 20. That is 2,025.This is the answer to the problem.

Let’s try 652:

1st step: The number 6
2nd step: 6 times 7 equals 42
3rd step: follow 42 with 25, which is 4225.

Algebraic proof:

Consider (ax + b)2 = a2. x2 +  2abx + b2.

This identity for  x = 10   and b = 5 becomes

(10a + 5) 2 a2 . 102 + 2. 10a . 5 + 52
a2 . 102 + a. 102 + 52
=  (a 2+ a ) . 102 + 52
a (a + 1) . 10 2 + 25

Clearly 10a + 5 represents two-digit numbers 15, 25, 35,…,95 for the values a = 1, 2, 3, … ,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25*.

Similar proof can be given for 3 or more digit numbers.

*The / symbol denotes that the numbers a(a+1) and 25 were concatenated.

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